Minimum Operations to Reduce X to Zero

🎯

Minimum Operations to Reduce X to Zero

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Imagine you have a budget X and a list of expenses. You want to remove expenses from either the start or end of the list to exactly match your budget, minimizing the number of removals.

💡 This problem asks you to find the minimum number of operations to reduce a number to zero by removing elements from either end of an array. Beginners often struggle because the problem looks like a two-pointer or DP problem, but the key insight is to track a complementary subarray sum using a hash map.

📋

Problem Statement

Given an integer array nums and an integer x, return the minimum number of operations to reduce x to exactly 0 by removing elements from the left or right end of the array. If it is not possible, return -1. Each operation consists of removing the leftmost or rightmost element from the array and subtracting its value from x.

1 ≤ nums.length ≤ 10^51 ≤ nums[i] ≤ 10^41 ≤ x ≤ 10^9

💡

Example

Input"nums = [1,1,4,2,3], x = 5"

Output2

Remove the last two elements (2 + 3) to reduce x to zero with 2 operations.

Input"nums = [5,6,7,8,9], x = 4"

Output-1

No combination of removals from ends can reduce x to zero.

Input"nums = [3,2,20,1,1,3], x = 10"

Output5

Remove the first three and last two elements (3+2+20+1+1+3) to reduce x to zero in 5 operations.

• x is exactly equal to sum of entire array → output is length of array
• x is smaller than any single element → output is -1
• nums contains only one element equal to x → output is 1
• nums contains multiple elements but no combination sums to x → output is -1

⚠️

Common Mistakes

Not considering the complement subarray sum and trying to remove from ends directly

Leads to inefficient brute force or incorrect logic

✅ Reframe problem to find longest subarray with sum totalSum - x

Overlapping prefix and suffix subarrays in hash map approach

Counting elements twice, leading to incorrect minimum operations

✅ Ensure suffix subarray starts after prefix subarray ends (suffix count > prefix count)

Not handling the case when x equals total sum of array

Returns -1 incorrectly or misses full array removal

✅ Check if target == 0 and return array length

Using sum() inside nested loops in brute force

Causes time limit exceeded due to O(n^3) complexity

✅ Precompute prefix sums or avoid brute force for large inputs

🧠

Brute Force (Try All Combinations from Both Ends)

💡 This approach exists to build intuition by exploring all possible ways to remove elements from the left and right ends. It is simple but inefficient, helping beginners understand the problem's constraints.

Intuition

Try every possible number of elements removed from the left, and for each, try removing elements from the right to see if the sum equals x.

Algorithm

1. For each possible count of elements removed from the left (0 to n):
2. Calculate the sum of these left elements.
3. For each possible count of elements removed from the right (0 to n - left_count):
4. Check if the sum of left and right elements equals x; if yes, update the minimum operations.

💡 The nested loops make it hard to see efficiency, but it clearly tries all combinations.

</>

Code

def minOperations(nums, x):
    n = len(nums)
    min_ops = float('inf')
    for left_count in range(n + 1):
        left_sum = sum(nums[:left_count])
        if left_sum > x:
            break
        for right_count in range(n - left_count + 1):
            right_sum = sum(nums[n - right_count:])
            if left_sum + right_sum == x:
                min_ops = min(min_ops, left_count + right_count)
    return min_ops if min_ops != float('inf') else -1

# Example usage
if __name__ == '__main__':
    print(minOperations([1,1,4,2,3], 5))  # Output: 2

Line Notes

for left_count in range(n + 1):Try all counts of elements removed from the left end

left_sum = sum(nums[:left_count])Calculate sum of left elements for current count

if left_sum > x:Early break to avoid unnecessary right side checks

for right_count in range(n - left_count + 1):Try all counts of elements removed from the right end given left_count

right_sum = sum(nums[n - right_count:])Calculate sum of right elements for current count

if left_sum + right_sum == x:Check if combined sum matches x

min_ops = min(min_ops, left_count + right_count)Update minimum operations if better

return min_ops if min_ops != float('inf') else -1Return result or -1 if no solution

import java.util.*;
public class Solution {
    public static int minOperations(int[] nums, int x) {
        int n = nums.length;
        int minOps = Integer.MAX_VALUE;
        for (int leftCount = 0; leftCount <= n; leftCount++) {
            int leftSum = 0;
            for (int i = 0; i < leftCount; i++) leftSum += nums[i];
            if (leftSum > x) break;
            for (int rightCount = 0; rightCount <= n - leftCount; rightCount++) {
                int rightSum = 0;
                for (int j = n - rightCount; j < n; j++) rightSum += nums[j];
                if (leftSum + rightSum == x) {
                    minOps = Math.min(minOps, leftCount + rightCount);
                }
            }
        }
        return minOps == Integer.MAX_VALUE ? -1 : minOps;
    }

    public static void main(String[] args) {
        System.out.println(minOperations(new int[]{1,1,4,2,3}, 5)); // Output: 2
    }
}

Line Notes

for (int leftCount = 0; leftCount <= n; leftCount++) {Try all counts of elements removed from left

for (int i = 0; i < leftCount; i++) leftSum += nums[i];Sum left elements for current count

if (leftSum > x) break;Stop early if left sum exceeds x

for (int rightCount = 0; rightCount <= n - leftCount; rightCount++) {Try all counts of elements removed from right

for (int j = n - rightCount; j < n; j++) rightSum += nums[j];Sum right elements for current count

if (leftSum + rightSum == x) {Check if combined sum equals x

minOps = Math.min(minOps, leftCount + rightCount);Update minimum operations

return minOps == Integer.MAX_VALUE ? -1 : minOps;Return result or -1 if no solution

#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int minOperations(vector<int>& nums, int x) {
    int n = nums.size();
    int minOps = INT_MAX;
    for (int leftCount = 0; leftCount <= n; leftCount++) {
        int leftSum = 0;
        for (int i = 0; i < leftCount; i++) leftSum += nums[i];
        if (leftSum > x) break;
        for (int rightCount = 0; rightCount <= n - leftCount; rightCount++) {
            int rightSum = 0;
            for (int j = n - rightCount; j < n; j++) rightSum += nums[j];
            if (leftSum + rightSum == x) {
                minOps = min(minOps, leftCount + rightCount);
            }
        }
    }
    return minOps == INT_MAX ? -1 : minOps;
}

int main() {
    vector<int> nums = {1,1,4,2,3};
    int x = 5;
    cout << minOperations(nums, x) << endl; // Output: 2
    return 0;
}

Line Notes

for (int leftCount = 0; leftCount <= n; leftCount++) {Try all counts of elements removed from left

for (int i = 0; i < leftCount; i++) leftSum += nums[i];Sum left elements for current count

if (leftSum > x) break;Early break to reduce unnecessary computation

for (int rightCount = 0; rightCount <= n - leftCount; rightCount++) {Try all counts of elements removed from right

for (int j = n - rightCount; j < n; j++) rightSum += nums[j];Sum right elements for current count

if (leftSum + rightSum == x) {Check if combined sum equals x

minOps = min(minOps, leftCount + rightCount);Update minimum operations

return minOps == INT_MAX ? -1 : minOps;Return result or -1 if no solution

function minOperations(nums, x) {
    const n = nums.length;
    let minOps = Infinity;
    for (let leftCount = 0; leftCount <= n; leftCount++) {
        let leftSum = 0;
        for (let i = 0; i < leftCount; i++) leftSum += nums[i];
        if (leftSum > x) break;
        for (let rightCount = 0; rightCount <= n - leftCount; rightCount++) {
            let rightSum = 0;
            for (let j = n - rightCount; j < n; j++) rightSum += nums[j];
            if (leftSum + rightSum === x) {
                minOps = Math.min(minOps, leftCount + rightCount);
            }
        }
    }
    return minOps === Infinity ? -1 : minOps;
}

// Example usage
console.log(minOperations([1,1,4,2,3], 5)); // Output: 2

Line Notes

for (let leftCount = 0; leftCount <= n; leftCount++) {Try all counts of elements removed from left

for (let i = 0; i < leftCount; i++) leftSum += nums[i];Sum left elements for current count

if (leftSum > x) break;Stop early if left sum exceeds x

for (let rightCount = 0; rightCount <= n - leftCount; rightCount++) {Try all counts of elements removed from right

for (let j = n - rightCount; j < n; j++) rightSum += nums[j];Sum right elements for current count

if (leftSum + rightSum === x) {Check if combined sum equals x

minOps = Math.min(minOps, leftCount + rightCount);Update minimum operations

return minOps === Infinity ? -1 : minOps;Return result or -1 if no solution

⏱

Complexity

TimeO(n^2)

SpaceO(1)

Two nested loops each up to n, and sum calls inside loops lead to O(n^2) time. Space is constant as no extra data structures used.

💡 For n=1000, this means about 1,000,000 operations, which is too slow for n=10^5.

Interview Verdict: TLE

This approach is too slow for large inputs but helps understand the problem and correctness before optimizing.

🧠

Prefix and Suffix Sum with Hash Map

💡 This approach improves efficiency by precomputing prefix sums and using a hash map to find complement sums quickly, reducing the problem to O(n) time.

Intuition

Compute prefix sums from the left and suffix sums from the right, store suffix sums in a hash map, then for each prefix sum, check if the complement (x - prefix_sum) exists in the suffix sums.

Algorithm

1. Compute suffix sums and store them in a hash map with their counts.
2. Iterate over prefix sums from the left, for each prefix sum, check if (x - prefix_sum) exists in the suffix sums map.
3. If found, update the minimum operations with sum of prefix and suffix counts.
4. Return the minimum operations or -1 if no solution.

💡 Using a hash map avoids nested loops by quickly finding complement sums.

</>

Code

def minOperations(nums, x):
    n = len(nums)
    suffix_sum_map = {0: 0}
    suffix_sum = 0
    for i in range(n - 1, -1, -1):
        suffix_sum += nums[i]
        suffix_sum_map[suffix_sum] = n - i
    min_ops = suffix_sum_map.get(x, float('inf'))
    prefix_sum = 0
    for i in range(n):
        prefix_sum += nums[i]
        if prefix_sum == x:
            min_ops = min(min_ops, i + 1)
        if (x - prefix_sum) in suffix_sum_map and suffix_sum_map[x - prefix_sum] > i + 1:
            min_ops = min(min_ops, i + 1 + suffix_sum_map[x - prefix_sum])
    return min_ops if min_ops != float('inf') else -1

# Example usage
if __name__ == '__main__':
    print(minOperations([1,1,4,2,3], 5))  # Output: 2

Line Notes

suffix_sum_map = {0: 0}Initialize map with zero sum for empty suffix

for i in range(n - 1, -1, -1):Compute suffix sums from right to left

suffix_sum_map[suffix_sum] = n - iStore suffix sum with count of elements removed

min_ops = suffix_sum_map.get(x, float('inf'))Check if suffix sum alone equals x

prefix_sum += nums[i]Accumulate prefix sum from left

if prefix_sum == x:Check if prefix sum alone equals x

if (x - prefix_sum) in suffix_sum_map and suffix_sum_map[x - prefix_sum] > i + 1:Check for complement suffix sum ensuring no overlap

return min_ops if min_ops != float('inf') else -1Return minimum operations or -1

import java.util.*;
public class Solution {
    public static int minOperations(int[] nums, int x) {
        int n = nums.length;
        Map<Integer, Integer> suffixSumMap = new HashMap<>();
        suffixSumMap.put(0, 0);
        int suffixSum = 0;
        for (int i = n - 1; i >= 0; i--) {
            suffixSum += nums[i];
            suffixSumMap.put(suffixSum, n - i);
        }
        int minOps = suffixSumMap.getOrDefault(x, Integer.MAX_VALUE);
        int prefixSum = 0;
        for (int i = 0; i < n; i++) {
            prefixSum += nums[i];
            if (prefixSum == x) minOps = Math.min(minOps, i + 1);
            int complement = x - prefixSum;
            if (suffixSumMap.containsKey(complement) && suffixSumMap.get(complement) > i + 1) {
                minOps = Math.min(minOps, i + 1 + suffixSumMap.get(complement));
            }
        }
        return minOps == Integer.MAX_VALUE ? -1 : minOps;
    }

    public static void main(String[] args) {
        System.out.println(minOperations(new int[]{1,1,4,2,3}, 5)); // Output: 2
    }
}

Line Notes

Map<Integer, Integer> suffixSumMap = new HashMap<>();Hash map to store suffix sums and counts

suffixSumMap.put(0, 0);Initialize with zero sum for empty suffix

for (int i = n - 1; i >= 0; i--) {Compute suffix sums from right

suffixSumMap.put(suffixSum, n - i);Map suffix sum to count of elements removed

int minOps = suffixSumMap.getOrDefault(x, Integer.MAX_VALUE);Check if suffix sum alone equals x

prefixSum += nums[i];Accumulate prefix sum from left

if (suffixSumMap.containsKey(complement) && suffixSumMap.get(complement) > i + 1) {Check complement suffix sum without overlap

return minOps == Integer.MAX_VALUE ? -1 : minOps;Return result or -1

#include <iostream>
#include <vector>
#include <unordered_map>
#include <climits>
using namespace std;

int minOperations(vector<int>& nums, int x) {
    int n = nums.size();
    unordered_map<int, int> suffixSumMap;
    suffixSumMap[0] = 0;
    int suffixSum = 0;
    for (int i = n - 1; i >= 0; i--) {
        suffixSum += nums[i];
        suffixSumMap[suffixSum] = n - i;
    }
    int minOps = suffixSumMap.count(x) ? suffixSumMap[x] : INT_MAX;
    int prefixSum = 0;
    for (int i = 0; i < n; i++) {
        prefixSum += nums[i];
        if (prefixSum == x) minOps = min(minOps, i + 1);
        int complement = x - prefixSum;
        if (suffixSumMap.count(complement) && suffixSumMap[complement] > i + 1) {
            minOps = min(minOps, i + 1 + suffixSumMap[complement]);
        }
    }
    return minOps == INT_MAX ? -1 : minOps;
}

int main() {
    vector<int> nums = {1,1,4,2,3};
    int x = 5;
    cout << minOperations(nums, x) << endl; // Output: 2
    return 0;
}

Line Notes

unordered_map<int, int> suffixSumMap;Hash map for suffix sums and counts

suffixSumMap[0] = 0;Initialize with zero sum for empty suffix

for (int i = n - 1; i >= 0; i--) {Compute suffix sums from right

suffixSumMap[suffixSum] = n - i;Map suffix sum to count of elements removed

int minOps = suffixSumMap.count(x) ? suffixSumMap[x] : INT_MAX;Check if suffix sum alone equals x

prefixSum += nums[i];Accumulate prefix sum from left

if (suffixSumMap.count(complement) && suffixSumMap[complement] > i + 1) {Check complement suffix sum without overlap

return minOps == INT_MAX ? -1 : minOps;Return result or -1

function minOperations(nums, x) {
    const n = nums.length;
    const suffixSumMap = new Map();
    suffixSumMap.set(0, 0);
    let suffixSum = 0;
    for (let i = n - 1; i >= 0; i--) {
        suffixSum += nums[i];
        suffixSumMap.set(suffixSum, n - i);
    }
    let minOps = suffixSumMap.has(x) ? suffixSumMap.get(x) : Infinity;
    let prefixSum = 0;
    for (let i = 0; i < n; i++) {
        prefixSum += nums[i];
        if (prefixSum === x) minOps = Math.min(minOps, i + 1);
        const complement = x - prefixSum;
        if (suffixSumMap.has(complement) && suffixSumMap.get(complement) > i + 1) {
            minOps = Math.min(minOps, i + 1 + suffixSumMap.get(complement));
        }
    }
    return minOps === Infinity ? -1 : minOps;
}

// Example usage
console.log(minOperations([1,1,4,2,3], 5)); // Output: 2

Line Notes

const suffixSumMap = new Map();Map to store suffix sums and counts

suffixSumMap.set(0, 0);Initialize with zero sum for empty suffix

for (let i = n - 1; i >= 0; i--) {Compute suffix sums from right

suffixSumMap.set(suffixSum, n - i);Map suffix sum to count of elements removed

let minOps = suffixSumMap.has(x) ? suffixSumMap.get(x) : Infinity;Check if suffix sum alone equals x

prefixSum += nums[i];Accumulate prefix sum from left

if (suffixSumMap.has(complement) && suffixSumMap.get(complement) > i + 1) {Check complement suffix sum without overlap

return minOps === Infinity ? -1 : minOps;Return result or -1

⏱

Complexity

TimeO(n)

SpaceO(n)

Single pass to build suffix sums map and single pass for prefix sums, each O(n). Space for hash map is O(n).

💡 For n=10^5, this approach is efficient enough to run within time limits.

Interview Verdict: Accepted

This approach is efficient and practical for large inputs, suitable for interviews.

🧠

Sliding Window to Find Longest Subarray with Sum = totalSum - x

💡 This approach reframes the problem to find the longest subarray whose sum equals totalSum - x, using a sliding window for O(n) time and O(1) space.

Intuition

Removing elements from the ends to reduce x to zero is equivalent to keeping a subarray in the middle with sum totalSum - x. Find the longest such subarray to minimize removals.

Algorithm

1. Calculate totalSum of the array.
2. Set target = totalSum - x.
3. Use two pointers (left, right) to find the longest subarray with sum equal to target.
4. Return n - length_of_longest_subarray or -1 if no such subarray exists.

💡 Sliding window efficiently finds the longest subarray with the target sum in linear time.

</>

Code

def minOperations(nums, x):
    totalSum = sum(nums)
    target = totalSum - x
    if target < 0:
        return -1
    if target == 0:
        return len(nums)
    left = 0
    current_sum = 0
    max_len = -1
    for right in range(len(nums)):
        current_sum += nums[right]
        while current_sum > target and left <= right:
            current_sum -= nums[left]
            left += 1
        if current_sum == target:
            max_len = max(max_len, right - left + 1)
    return len(nums) - max_len if max_len != -1 else -1

# Example usage
if __name__ == '__main__':
    print(minOperations([1,1,4,2,3], 5))  # Output: 2

Line Notes

totalSum = sum(nums)Calculate total sum of array

target = totalSum - xFind target sum for subarray to keep

if target < 0:If target negative, no solution

left = 0Initialize left pointer for sliding window

current_sum += nums[right]Expand window by adding right element

while current_sum > target and left <= right:Shrink window from left if sum too large

if current_sum == target:Check if current window matches target sum

return len(nums) - max_len if max_len != -1 else -1Calculate minimum removals or return -1

public class Solution {
    public static int minOperations(int[] nums, int x) {
        int totalSum = 0;
        for (int num : nums) totalSum += num;
        int target = totalSum - x;
        if (target < 0) return -1;
        if (target == 0) return nums.length;
        int left = 0, currentSum = 0, maxLen = -1;
        for (int right = 0; right < nums.length; right++) {
            currentSum += nums[right];
            while (currentSum > target && left <= right) {
                currentSum -= nums[left];
                left++;
            }
            if (currentSum == target) {
                maxLen = Math.max(maxLen, right - left + 1);
            }
        }
        return maxLen == -1 ? -1 : nums.length - maxLen;
    }

    public static void main(String[] args) {
        System.out.println(minOperations(new int[]{1,1,4,2,3}, 5)); // Output: 2
    }
}

Line Notes

int totalSum = 0;Calculate total sum of array

int target = totalSum - x;Target sum for subarray to keep

if (target < 0) return -1;No solution if target negative

int left = 0, currentSum = 0, maxLen = -1;Initialize sliding window pointers and max length

currentSum += nums[right];Expand window by adding right element

while (currentSum > target && left <= right) {Shrink window from left if sum too large

if (currentSum == target) {Check if current window matches target

return maxLen == -1 ? -1 : nums.length - maxLen;Return minimum removals or -1

#include <iostream>
#include <vector>
using namespace std;

int minOperations(vector<int>& nums, int x) {
    int totalSum = 0;
    for (int num : nums) totalSum += num;
    int target = totalSum - x;
    if (target < 0) return -1;
    if (target == 0) return nums.size();
    int left = 0, currentSum = 0, maxLen = -1;
    for (int right = 0; right < (int)nums.size(); right++) {
        currentSum += nums[right];
        while (currentSum > target && left <= right) {
            currentSum -= nums[left];
            left++;
        }
        if (currentSum == target) {
            maxLen = max(maxLen, right - left + 1);
        }
    }
    return maxLen == -1 ? -1 : (int)nums.size() - maxLen;
}

int main() {
    vector<int> nums = {1,1,4,2,3};
    int x = 5;
    cout << minOperations(nums, x) << endl; // Output: 2
    return 0;
}

Line Notes

int totalSum = 0;Calculate total sum of array

int target = totalSum - x;Target sum for subarray to keep

if (target < 0) return -1;No solution if target negative

int left = 0, currentSum = 0, maxLen = -1;Initialize sliding window pointers and max length

currentSum += nums[right];Expand window by adding right element

while (currentSum > target && left <= right) {Shrink window from left if sum too large

if (currentSum == target) {Check if current window matches target

return maxLen == -1 ? -1 : (int)nums.size() - maxLen;Return minimum removals or -1

function minOperations(nums, x) {
    const totalSum = nums.reduce((a, b) => a + b, 0);
    const target = totalSum - x;
    if (target < 0) return -1;
    if (target === 0) return nums.length;
    let left = 0, currentSum = 0, maxLen = -1;
    for (let right = 0; right < nums.length; right++) {
        currentSum += nums[right];
        while (currentSum > target && left <= right) {
            currentSum -= nums[left];
            left++;
        }
        if (currentSum === target) {
            maxLen = Math.max(maxLen, right - left + 1);
        }
    }
    return maxLen === -1 ? -1 : nums.length - maxLen;
}

// Example usage
console.log(minOperations([1,1,4,2,3], 5)); // Output: 2

Line Notes

const totalSum = nums.reduce((a, b) => a + b, 0);Calculate total sum of array

const target = totalSum - x;Target sum for subarray to keep

if (target < 0) return -1;No solution if target negative

let left = 0, currentSum = 0, maxLen = -1;Initialize sliding window pointers and max length

currentSum += nums[right];Expand window by adding right element

while (currentSum > target && left <= right) {Shrink window from left if sum too large

if (currentSum === target) {Check if current window matches target

return maxLen === -1 ? -1 : nums.length - maxLen;Return minimum removals or -1

⏱

Complexity

TimeO(n)

SpaceO(1)

Single pass sliding window with two pointers, constant extra space.

💡 This is the most efficient approach, suitable for very large inputs.

Interview Verdict: Accepted

Optimal solution with minimal space and linear time, ideal for interviews.

📊

All Approaches - One-Glance Tradeoffs

💡 Use the sliding window approach in 95% of interviews due to its optimal time and space complexity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n^2)	O(1)	No	Yes	Mention only - never code
2. Prefix and Suffix Sum with Hash Map	O(n)	O(n)	No	Yes	Good to mention as optimization
3. Sliding Window	O(n)	O(1)	No	Yes	Code this approach

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding all approaches, and prepare to explain tradeoffs clearly in interviews.

How to Present

Step 1: Clarify the problem and constraints with the interviewer.Step 2: Describe the brute force approach to show understanding.Step 3: Explain the prefix-suffix sum with hash map optimization.Step 4: Present the sliding window approach as the optimal solution.Step 5: Code the sliding window solution and test with examples.

Time Allocation

Clarify: 2min → Approach discussion: 5min → Code: 10min → Testing & edge cases: 3min. Total ~20min

What the Interviewer Tests

The interviewer tests your ability to recognize complementary subarray sums, optimize brute force, and implement sliding window correctly.

Common Follow-ups

• What if elements can be removed from anywhere, not just ends? → This becomes a different problem, possibly subset sum.
• How to handle very large x values efficiently? → Use prefix sums and early pruning.
• Can you reconstruct the actual sequence of removals? → Yes, by tracking indices during sliding window.
• What if negative numbers are allowed? → Sliding window may not work; consider hashmap with prefix sums.

💡 These follow-ups test your understanding of problem variations and edge cases.

🔍

Pattern Recognition

When to Use

1) Problem involves removing elements from ends; 2) Target sum to reduce; 3) Need minimum operations; 4) Array with positive integers

Signature Phrases

remove elements from left or rightreduce x to zerominimum number of operations

NOT This Pattern When

Problems that allow removing elements from anywhere or involve negative numbers require different approaches.

Similar Problems

Two Sum - uses hash map for complement sumsSubarray Sum Equals K - finding subarrays with target sumLongest Subarray with Sum K - sliding window technique

Practice

1. Given an array of integers and an integer k, you need to determine if the array can be partitioned into pairs such that the sum of each pair is divisible by k. Which approach guarantees an optimal solution with linear time complexity?

easy

A. Use a frequency map to count remainders modulo k and verify complementary counts.

B. Check all pairs with nested loops to find valid pairs.

C. Use dynamic programming to find subsets whose sums are divisible by k.

D. Use a greedy approach by pairing elements with the smallest remainder first.

Solution

1. Step 1: Understand the problem constraints

   The problem requires pairing elements so that each pair sums to a multiple of k. This depends on the remainders of elements modulo k.

2. Step 2: Identify the approach that efficiently tracks remainder frequencies

   Using a frequency map of remainders allows pairing elements with complementary remainders (e.g., remainder r pairs with k-r). This approach runs in O(n) time and uses O(k) space.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Frequency map approach matches the optimal solution pattern [OK]

Hint: Frequency map of remainders solves pairing optimally [OK]

Common Mistakes:

• Assuming greedy pairing by smallest remainder always works
• Using DP which is unnecessary and inefficient here
• Brute force nested loops for large inputs

2. You are given an array of integers and need to find an index where the sum of elements to the left equals the sum of elements to the right. Which approach guarantees an optimal O(n) time and O(1) space solution?

easy

A. Maintain a running left sum and use the total sum to check the right sum in a single pass.

B. Use dynamic programming to store prefix sums and then binary search for the pivot index.

C. Use a brute force nested loop to compute left and right sums for each index.

D. Sort the array and use two pointers to find the pivot index.

Solution

1. Step 1: Understand the problem constraints

   The problem requires finding an index where left sum equals right sum efficiently.

2. Step 2: Identify the optimal approach

   Maintaining a running left sum and using the total sum to compute right sum on the fly allows a single pass with O(1) extra space.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Single pass with running sums is O(n) time and O(1) space [OK]

Hint: Running left sum + total sum enables single pass [OK]

Common Mistakes:

• Assuming nested loops are necessary
• Thinking sorting helps with sums
• Confusing prefix sums with binary search

3. You are given a sorted list of unique integers and need to summarize consecutive sequences as ranges (e.g., [0,1,2,4,5,7] -> ['0->2','4->5','7']). Which approach guarantees an optimal O(n) time solution with minimal extra space?

easy

A. Using two pointers to scan the list once, expanding the end pointer while consecutive numbers continue, then recording ranges.

B. Using nested loops to check every possible range start and end, appending ranges when sequences break.

C. Using dynamic programming to store the longest consecutive subsequence ending at each index and reconstruct ranges.

D. Using a greedy approach that picks the next number if it is consecutive, otherwise starts a new range, but restarts scanning from the beginning each time.

Solution

1. Step 1: Understand the problem constraints

   The input is sorted and unique, so consecutive sequences can be detected by checking if nums[end + 1] == nums[end] + 1.

2. Step 2: Identify the optimal approach

   Two pointers allow a single pass: one pointer marks the start of a range, the other expands until the sequence breaks, then record the range and move start pointer forward.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Two pointers scan once -> O(n) time and O(n) space for output [OK]

Hint: Two pointers scan sorted list once for consecutive ranges [OK]

Common Mistakes:

• Thinking nested loops are necessary
• Confusing DP with this linear scan
• Restarting scan repeatedly wastes time

4. Suppose the problem changes: now you want the longest substring with at most two distinct characters, but characters can be reused infinitely many times and the input string is extremely long. Which modification to the sliding window approach is necessary to handle this efficiently?

hard

A. Maintain a sliding window with frequency counts but optimize by using a fixed-size array for frequencies instead of a hash map.

B. Use a brute force nested loop approach with early breaks to handle reuse.

C. Use a priority queue to track the least recently used character to evict when distinct count exceeds two.

D. Switch to a dynamic programming approach that stores results for substrings with reuse allowed.

Solution

1. Step 1: Understand reuse and input size impact

   Characters can be reused infinitely, but the problem remains longest substring with at most two distinct chars. The input is very long, so efficiency is critical.

2. Step 2: Identify efficient data structure

   Using a fixed-size array for frequencies (size 128 for ASCII) is more efficient than a hash map, reducing overhead and ensuring O(n) time and O(1) space.

3. Step 3: Confirm approach correctness

   The sliding window logic remains the same, but frequency tracking is optimized for speed and memory, suitable for very long inputs with reuse.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Fixed-size array frequency tracking is optimal for large inputs with limited charset [OK]

Hint: Fixed-size array beats hash map for fixed charset [OK]

Common Mistakes:

• Using brute force despite input size
• Using complex data structures unnecessarily
• Switching to DP without need

5. Suppose the input array can contain multiple repeated elements and you want to find the longest subarray with equal number of 0s and 1s, but now elements can be reused multiple times (like a circular array). Which modification to the prefix sum approach correctly handles this scenario?

hard

A. Use a sliding window with two pointers to find balanced subarrays in the circular array.

B. Reset prefix sums and first_occurrence after each full pass to handle reuse.

C. Extend the array by concatenating it to itself and run the prefix sum algorithm on the doubled array.

D. Modify prefix sums to be modulo n and track first occurrences modulo n.

Solution

1. Step 1: Recognize circular array means subarrays can wrap around.

   Doubling the array simulates wrap-around by linearizing circular behavior.

2. Step 2: Run prefix sum algorithm on doubled array to find longest balanced subarray crossing boundary.

   This preserves O(n) complexity and correctness.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Other options break correctness or do not handle wrap-around properly [OK]

Hint: Double array to handle circular subarrays [OK]

Common Mistakes:

• Resetting prefix sums loses info
• Modulo prefix sums break balance detection
• Sliding window can't detect all wrap-around cases