Binary Search on Answer Technique in DSA Go

Mental Model

We guess an answer and check if it works, then adjust the guess up or down until we find the best answer.

Analogy: Imagine trying to find the right temperature to bake a cake by guessing and adjusting based on whether it's undercooked or burnt.

low
 ↑
[? ? ? ? ? ? ? ? ? ?]
                      ↑
high

We pick a middle guess between low and high and test it.

Dry Run Walkthrough

Input: array: [1, 2, 3, 4, 5], target sum: 8, find max length of subarray with sum <= 8

Goal: Find the longest subarray length whose sum is at most 8 using binary search on answer

Step 1: Set low=1, high=5 (array length), mid=(1+5)/2=3; check if subarray length 3 can have sum <= 8

array: [1, 2, 3, 4, 5]
Trying length=3
Subarrays: [1,2,3]=6 ≤ 8 (valid)
[2,3,4]=9 > 8 (invalid)
[3,4,5]=12 > 8 (invalid)

Why: We test mid length to see if any subarray of this length meets the sum condition

Step 2: Since length 3 is valid (some subarray sum ≤ 8), set low=mid+1=4 to try longer subarrays

low=4, high=5
Trying length=4

Why: Try to find a longer valid subarray

Step 3: Check subarrays of length 4: [1,2,3,4]=10 > 8 (invalid), [2,3,4,5]=14 > 8 (invalid)

No valid subarray of length 4

Why: Length 4 is too long, sum exceeds target

Step 4: Since length 4 invalid, set high=mid-1=3

low=4, high=3 (low > high, stop)

Why: No longer subarray possible, binary search ends

Result:

Longest valid subarray length = 3

Annotated Code

DSA Go

package main

import "fmt"

func canFindSubarray(arr []int, length int, target int) bool {
    sum := 0
    for i := 0; i < length; i++ {
        sum += arr[i] // sum first 'length' elements
    }
    if sum <= target {
        return true
    }
    for i := length; i < len(arr); i++ {
        sum += arr[i] - arr[i-length] // slide window
        if sum <= target {
            return true
        }
    }
    return false
}

func binarySearchOnAnswer(arr []int, target int) int {
    low, high := 1, len(arr)
    result := 0
    for low <= high {
        mid := (low + high) / 2
        if canFindSubarray(arr, mid, target) {
            result = mid // mid length works, try longer
            low = mid + 1
        } else {
            high = mid - 1 // mid length too long, try shorter
        }
    }
    return result
}

func main() {
    arr := []int{1, 2, 3, 4, 5}
    target := 8
    fmt.Printf("Longest valid subarray length = %d\n", binarySearchOnAnswer(arr, target))
}

for low <= high {

binary search loop to narrow down answer range

mid := (low + high) / 2

guess middle length to test

if canFindSubarray(arr, mid, target) {

check if subarray of length mid meets sum condition

result = mid // mid length works, try longer

record valid length and try longer

low = mid + 1

increase lower bound to search longer lengths

high = mid - 1 // mid length too long, try shorter

decrease upper bound to search shorter lengths

sum += arr[i] - arr[i-length] // slide window

update sum for next subarray by sliding window

Output

Longest valid subarray length = 3

Complexity Analysis

Time: O(n log n) because each binary search step does a sliding window check over n elements

Space: O(1) because only a few variables and counters are used

vs Alternative: Naive approach checks all subarrays O(n^2), binary search on answer reduces to O(n log n)

Edge Cases

empty array

returns 0 because no subarray exists

DSA Go

low, high := 1, len(arr) // high=0 means loop never runs

target smaller than smallest element

returns 0 because no subarray sum can be ≤ target

DSA Go

if canFindSubarray(arr, mid, target) { ... } else { ... }

all elements equal and sum exactly target

returns max length fitting sum exactly

DSA Go

sum <= target check in canFindSubarray

When to Use This Pattern

When you need to find the best numeric answer but can't check all answers directly, use binary search on answer to guess and verify efficiently.

Common Mistakes

Mistake: Not updating low and high correctly after checking mid
Fix: Set low = mid + 1 when mid is valid, else high = mid - 1

Mistake: Checking subarray sums inefficiently inside binary search
Fix: Use sliding window to check sums in O(n) per guess

Summary

It finds the best answer by guessing and checking with binary search.

Use it when the answer is a number and you can verify guesses efficiently.

The key is to turn the problem into yes/no questions about guesses and narrow down.

Practice

1. What is the main idea behind the Binary Search on Answer technique?

easy

A. We guess an answer and check if it satisfies the problem, then adjust the guess using binary search.

B. We sort the input array before applying binary search.

C. We use binary search to find an element in a sorted array.

D. We use recursion to divide the problem into smaller subproblems.

Solution

1. Step 1: Understand the technique concept

   Binary Search on Answer guesses a possible answer and checks if it meets the problem's conditions.

2. Step 2: Adjust guess based on check

   If the guess is too high or too low, we adjust it using binary search until the best answer is found.

3. Final Answer:

   We guess an answer and check if it satisfies the problem, then adjust the guess using binary search. -> Option A

4. Quick Check:

   Guess and check with binary search = D [OK]

Hint: Think guess-check-adjust using binary search [OK]

Common Mistakes:

• Confusing with normal binary search on arrays
• Thinking it always requires sorting input
• Assuming recursion is mandatory

2. Which of the following is the correct Go syntax to start a binary search on answer between low = 1 and high = 100?

easy

A. while low < high { mid = (low + high) / 2 }

B. for i := low; i < high; i++ { mid := (low + high) / 2 }

C. for low <= high { mid := (low + high) / 2 }

D. if low < high { mid := (low + high) / 2 }

Solution

1. Step 1: Identify Go loop syntax for binary search

   Go uses for loops for iteration; for low <= high { ... } is correct for binary search.

2. Step 2: Check other options for syntax errors

   while is not valid in Go; for i := low; i < high; i++ is a counting loop, not binary search; if is conditional, not loop.

3. Final Answer:

   for low <= high { mid := (low + high) / 2 } -> Option C

4. Quick Check:

   Go binary search loop uses for low <= high [OK]

Hint: Use for loop with condition low <= high in Go [OK]

Common Mistakes:

• Using while loop (not valid in Go)
• Using for with increment instead of condition
• Using if instead of loop

3. Consider the following Go code snippet using binary search on answer to find the minimum x such that x*x >= 10:

low, high := 1, 10
var ans int
for low <= high {
  mid := (low + high) / 2
  if mid*mid >= 10 {
    ans = mid
    high = mid - 1
  } else {
    low = mid + 1
  }
}
fmt.Println(ans)

What will be printed?

medium

A. 3

B. 4

C. 10

D. 1

Solution

1. Step 1: Trace binary search for x where x*x >= 10

   Check mid values: - mid=5: 25 >=10 true, ans=5, high=4 - mid=2: 4 >=10 false, low=3 - mid=3: 9 >=10 false, low=4 - mid=4: 16 >=10 true, ans=4, high=3 Loop ends, ans=4

2. Step 2: Find minimum x satisfying condition

   Minimum x with x*x >=10 is 4, so ans=4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Minimum x with x²≥10 is 4 [OK]

Hint: Check mid² against target, adjust low/high accordingly [OK]

Common Mistakes:

• Choosing mid incorrectly (integer division errors)
• Not updating ans when condition is true
• Confusing >= with >

4. The following Go code tries to find the maximum length of a rope that can be cut into at least k pieces using binary search on answer. Identify the bug:

func maxRopeLength(ropes []int, k int) int {
  low, high := 1, 0
  for _, r := range ropes {
    if r > high {
      high = r
    }
  }
  ans := 0
  for low < high {
    mid := (low + high) / 2
    count := 0
    for _, r := range ropes {
      count += r / mid
    }
    if count >= k {
      ans = mid
      low = mid + 1
    } else {
      high = mid - 1
    }
  }
  return ans
}

medium

A. The loop condition should be low <= high, not low < high.

B. The initial high value should be 1, not 0.

C. The count calculation should use float division, not integer division.

D. The ans variable should be initialized to high, not 0.

Solution

1. Step 1: Analyze binary search loop condition

   The loop uses for low < high, which misses the case when low == high, possibly skipping the final answer.

2. Step 2: Correct loop condition for binary search

   Changing to for low <= high ensures all candidates are checked, including when low equals high.

3. Final Answer:

   The loop condition should be low <= high, not low < high. -> Option A

4. Quick Check:

   Binary search loops use low <= high [OK]

Hint: Use low <= high to include all candidates in binary search [OK]

Common Mistakes:

• Using low < high and missing final check
• Confusing integer and float division here
• Initializing ans incorrectly

5. You want to find the minimum maximum workload to assign n tasks to m workers so that no worker has workload exceeding that value. You have an array of task times. Which approach correctly applies binary search on answer to solve this problem?

hard

A. Use binary search on the number of workers instead of workload.

B. Set low as 0, high as max task time; binary search on this range checking if sum of tasks is less than high.

C. Sort tasks and binary search for the task with median time as answer.

D. Set low as max task time, high as sum of all tasks; binary search on this range checking feasibility of workload limit.

Solution

1. Step 1: Define search space for answer

   The minimum maximum workload must be at least the largest single task time (low) and at most the sum of all tasks (high).

2. Step 2: Check feasibility in binary search

   For each mid in [low, high], check if tasks can be assigned to m workers without exceeding mid workload per worker.

3. Step 3: Adjust search based on feasibility

   If feasible, try smaller max workload (high = mid - 1), else increase (low = mid + 1).

4. Final Answer:

   Set low as max task time, high as sum of all tasks; binary search on this range checking feasibility of workload limit. -> Option D

5. Quick Check:

   Search between max task and sum, check feasibility [OK]

Hint: Search between max single task and total sum [OK]

Common Mistakes:

• Setting low to 0 ignoring max task time
• Searching on wrong variable like number of workers
• Not checking feasibility properly