Largest Subarray with Zero Sum code example

Example 1: How to find the suarray with maximum sum using divide and conquer

#include <stdio.h>
#include <limits.h>
 
// Utility function to find maximum of two numbers
int max(int x, int y) {
    return (x > y) ? x : y;
}
 
// Function to find maximum subarray sum using divide and conquer
int maximum_sum(int A[], int low, int high)
{
    // If array contains only one element
    if (high == low)
        return A[low];
 
    // Find middle element of the array
    int mid = (low + high) / 2;
 
    // Find maximum subarray sum for the left subarray
    // including the middle element
    int left_max = INT_MIN;
    int sum = 0;
    for (int i = mid; i >= low; i--)
    {
        sum += A[i];
        if (sum > left_max)
            left_max = sum;
    }
 
    // Find maximum subarray sum for the right subarray
    // excluding the middle element
    int right_max = INT_MIN;
    sum = 0;    // reset sum to 0
    for (int i = mid + 1; i <= high; i++)
    {
        sum += A[i];
        if (sum > right_max)
            right_max = sum;
    }
 
    // Recursively find the maximum subarray sum for left subarray
    // and right subarray and take maximum
    int max_left_right = max(maximum_sum(A, low, mid),
                            maximum_sum(A, mid + 1, high));
 
    // return maximum of the three
    return max(max_left_right, left_max + right_max);
}
 
// Maximum Sum Subarray using Divide & Conquer
int main(void)
{
    int arr[] = { 2, -4, 1, 9, -6, 7, -3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    printf("The maximum sum of the subarray is %d", 
            maximum_sum(arr, 0, n - 1));
 
    return 0;
}

Example 2: largest subarray of 0's and 1's

public class Solution {

    public int findMaxLength(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int maxlen = 0, count = 0;
        for (int i = 0; i < nums.length; i++) {
            count = count + (nums[i] == 1 ? 1 : -1);
            if (map.containsKey(count)) {
                maxlen = Math.max(maxlen, i - map.get(count));
            } else {
                map.put(count, i);
            }
        }
        return maxlen;
    }
}