Partition Equal Subset Sum in DSA C - Build from Scratch

Partition Equal Subset Sum

📖 Scenario: You have a collection of items with weights represented as integers. You want to check if it is possible to split these items into two groups so that both groups have the same total weight.This is useful in real life when you want to pack two balanced boxes or divide tasks evenly.

🎯 Goal: Build a program that determines if the given list of integers can be partitioned into two subsets with equal sum.

📋 What You'll Learn

Create an integer array called nums with the exact values: 1, 5, 11, 5

Create an integer variable called total_sum to hold the sum of all elements in nums

Implement a function called canPartition that returns 1 if the array can be partitioned into two subsets with equal sum, otherwise 0

Print "YES" if partition is possible, else print "NO"

💡 Why This Matters

🌍 Real World

Balancing loads, dividing tasks or resources evenly, and packing problems often require checking if a set can be split into equal parts.

💼 Career

Understanding subset sum and partition problems is important in software engineering roles involving algorithms, optimization, and problem solving.

1

Create the array of numbers

Create an integer array called nums with the exact values 1, 5, 11, 5 and an integer variable n that stores the number of elements in nums.

DSA C

int nums[] = {1, 5, 11, 5};
int n = sizeof(nums) / sizeof(nums[0]);
// Your code here

Hint

Use sizeof(nums) / sizeof(nums[0]) to get the number of elements in the array.

2

Calculate the total sum of the array

Create an integer variable called total_sum and set it to 0. Then use a for loop with variable i from 0 to n - 1 to add each element of nums to total_sum.

DSA C

int nums[] = {1, 5, 11, 5};
int n = sizeof(nums) / sizeof(nums[0]);
int total_sum = 0;
// Your code here

Hint

Use a for loop to add each number in nums to total_sum.

3

Implement the partition check function

Write a function called canPartition that takes nums and n as parameters. Inside, if total_sum is odd, return 0. Otherwise, use a boolean array dp of size total_sum / 2 + 1 initialized to 0 except dp[0] set to 1. Use nested loops to update dp to check if a subset sum equal to total_sum / 2 is possible. Return dp[total_sum / 2].

DSA C

int nums[] = {1, 5, 11, 5};
int n = sizeof(nums) / sizeof(nums[0]);
int total_sum = 0;
for (int i = 0; i < n; i++) {
    total_sum += nums[i];
}

int canPartition(int* nums, int n) {
    // Your code here
}

Hint

Use dynamic programming with a boolean array to track possible subset sums.

4

Print the result

Use printf to print "YES" if canPartition(nums, n) returns 1, otherwise print "NO".

DSA C

int nums[] = {1, 5, 11, 5};
int n = sizeof(nums) / sizeof(nums[0]);
int total_sum = 0;
for (int i = 0; i = nums[i]; j--) {
            if (dp[j - nums[i]] == 1) {
                dp[j] = 1;
            }
        }
    }
    return dp[half];
}

// Your code here

Hint

Use printf("YES\n") or printf("NO\n") based on the function result.

Practice

1. What does the Partition Equal Subset Sum problem ask you to find?

easy

A. If the array contains any duplicate elements

B. If the array is sorted in ascending order

C. If the array can be split into two subsets with equal sums

D. If the array contains only positive numbers

Solution

1. Step 1: Understand the problem statement

   The problem asks whether the given array can be divided into two parts where the sum of elements in both parts is the same.

2. Step 2: Identify the key requirement

   We need to check if there exists a subset whose sum is half of the total sum of the array.

3. Final Answer:

   If the array can be split into two subsets with equal sums -> Option C

4. Quick Check:

   Partition Equal Subset Sum = Split array equally [OK]

Hint: Check if total sum is even first to proceed [OK]

Common Mistakes:

• Thinking the problem is about sorting
• Checking for duplicates instead of sums
• Assuming all numbers must be positive

2. Which of the following is the correct way to declare a boolean array in C for dynamic programming in Partition Equal Subset Sum?

easy

A. boolean dp[sum+1];

B. int dp[sum+1];

C. bool dp = new bool[sum+1];

D. bool dp[sum+1];

Solution

1. Step 1: Identify correct C syntax for boolean arrays

   In C, bool is available via stdbool.h and arrays are declared as bool dp[size];.

2. Step 2: Check options for correctness

   bool dp[sum+1]; uses correct syntax. int dp[sum+1]; uses int, which is not boolean. bool dp = new bool[sum+1]; uses Java/C++ style dynamic allocation, not valid in C. boolean dp[sum+1]; uses 'boolean' which is not a C keyword.

3. Final Answer:

   bool dp[sum+1]; -> Option D

4. Quick Check:

   Use bool dp[size]; in C [OK]

Hint: Use stdbool.h and declare bool arrays with bool dp[size]; [OK]

Common Mistakes:

• Using 'boolean' instead of 'bool' in C
• Trying Java-style dynamic allocation in C
• Using int instead of bool for DP array

3. What will be the output of the following code snippet for array nums = {1, 5, 11, 5}?

#include 
#include 

bool canPartition(int* nums, int numsSize) {
    int sum = 0;
    for (int i = 0; i < numsSize; i++) sum += nums[i];
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    bool dp[target + 1];
    for (int i = 0; i <= target; i++) dp[i] = false;
    dp[0] = true;
    for (int i = 0; i < numsSize; i++) {
        for (int j = target; j >= nums[i]; j--) {
            dp[j] = dp[j] || dp[j - nums[i]];
        }
    }
    return dp[target];
}

int main() {
    int nums[] = {1, 5, 11, 5};
    printf("%d\n", canPartition(nums, 4));
    return 0;
}

medium

A. 1

B. 0

C. Compilation error

D. Runtime error

Solution

1. Step 1: Calculate total sum and check parity

   Sum = 1 + 5 + 11 + 5 = 22, which is even, so proceed.

2. Step 2: Use DP to check subset sum equals 11

   The DP array tracks sums achievable. The subset {11} and {1,5,5} both sum to 11, so dp[11] becomes true.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Subset sum exists = 1 [OK]

Hint: Check if half sum subset exists using DP [OK]

Common Mistakes:

• Not checking if total sum is even
• Incorrect DP initialization
• Looping forward instead of backward in DP

4. Identify the bug in this Partition Equal Subset Sum code snippet:

bool canPartition(int* nums, int numsSize) {
    int sum = 0;
    for (int i = 0; i < numsSize; i++) sum += nums[i];
    if (sum % 2 != 0) return false;
    int target = sum / 2;
    bool dp[target + 1];
    for (int i = 0; i < target; i++) dp[i] = false;
    dp[0] = true;
    for (int i = 0; i < numsSize; i++) {
        for (int j = target; j >= nums[i]; j--) {
            dp[j] = dp[j] || dp[j - nums[i]];
        }
    }
    return dp[target];
}

medium

A. Loop initializing dp should run till target inclusive

B. The sum calculation is incorrect

C. The inner loop should run forward, not backward

D. dp[0] should be false initially

Solution

1. Step 1: Check dp array initialization loop

   The loop runs from 0 to target-1, missing dp[target] initialization, which should be set to false.

2. Step 2: Correct the loop to include target index

   Change loop condition to i <= target to initialize all dp elements properly.

3. Final Answer:

   Loop initializing dp should run till target inclusive -> Option A

4. Quick Check:

   Initialize dp array fully including dp[target] [OK]

Hint: Initialize dp array fully including last index [OK]

Common Mistakes:

• Forgetting to initialize dp[target]
• Running inner loop forward causing wrong results
• Setting dp[0] to false instead of true

5. Given an array nums = {2, 2, 3, 5}, which of the following modifications will correctly determine if it can be partitioned into two subsets with equal sum using dynamic programming?

hard

A. Change the DP array to track counts of subsets instead of booleans

B. Use a 2D DP array where dp[i][j] means if sum j is possible with first i elements

C. Sort the array first and then apply the existing 1D DP approach

D. Only check if the largest element is less than half the total sum

Solution

1. Step 1: Understand limitations of 1D DP for some cases

   1D DP works but 2D DP is clearer and handles subset sums by tracking elements used.

2. Step 2: Use 2D DP where dp[i][j] shows if sum j is possible using first i elements

   This approach explicitly considers each element and sum, making it easier to debug and understand.

3. Step 3: Sorting or counting subsets is not necessary or sufficient

   Sorting doesn't affect subset sums. Counting subsets is more complex and not required for yes/no answer.

4. Final Answer:

   Use a 2D DP array where dp[i][j] means if sum j is possible with first i elements -> Option B

5. Quick Check:

   2D DP tracks sums per element [OK]

Hint: Use 2D DP for clear subset sum tracking [OK]

Common Mistakes:

• Sorting array expecting it to solve partition
• Counting subsets instead of checking existence
• Only checking largest element against half sum