Explanation:

• Calculate prefix sum with give that leftSum = 0 so prefixSum[0] = 0
• Calculate suffix sum with given that right sum = 0 so suffixSum[n-1] = 0
• Calculate the difference between prefixSum and suffixSum array
• time and space complexity O(n)

public int[] findDifferenceArray(int[] nums) {
        int n = nums.length;
        int[] differenceArray = new int[n];
        int[] prefixSum = new int[n];
        int[] suffixSum = new int[n];
        prefixSum[0] = 0;
        suffixSum[n-1] = 0;
        for(int i = 1; i< n; i++) {
            prefixSum[i] = prefixSum[i-1] + nums[i-1]; 
        }

        
        for(int i = n-2; i>= 0; i--) {
            suffixSum[i] = suffixSum[i+1] + nums[i+1]; 
        }
        
        for(int i =0;i<n;i++) {
            differenceArray[i] = Math.abs(prefixSum[i] - suffixSum[i]);
        }

        return differenceArray;
    }