Explanation:

• Calculate prefix sum and suffix sum
• check where both the prefix sum and suffix sum are equals that would be our resulting index and break the loop
• if not matches then resulting index is -1
• time complexity and space complexity both are O(n)

// Method to find the index where the sum of elements to the left equals the sum of elements to the right
    public int findMiddleIndex(int[] nums) {
        
        int[] prefixSum = new int[nums.length];
        int[] suffixSum = new int[nums.length];
        prefixSum[0] = 0;
        suffixSum[nums.length-1] = 0;

        for(int i =1; i< nums.length; i++) {
            prefixSum[i] = prefixSum[i-1] + nums[i-1];
        }

        for(int i= nums.length-2;i>=0;i--) {
            suffixSum[i] = suffixSum[i+1] + nums[i+1];
        }

        int outputIndex = -1;
        for(int i =0; i< nums.length; i++) {
            if(prefixSum[i] == suffixSum[i]) {
                outputIndex = i;
                break;
            }
        }


        return outputIndex; 
    }