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Sherlock and Array

  • emhhacker
     + 0 comments

    My Java solution with O(n) time complexity and O(1) space complexity:

    public static String balancedSums(List<Integer> arr) {
        // goal: find element of arr where sum of elements to left equals sum of elements to the right
        if(arr.size() == 1) return "YES"; //left = 0 = right
        
        int totalSum = arr.stream().mapToInt(Integer::intValue).sum();
        int leftSum = 0;

        for (int i = 0; i < arr.size(); i++) {
            int rightSum = totalSum - leftSum - arr.get(i);
            if (leftSum == rightSum) return "YES";
            leftSum += arr.get(i);
        }
        return "NO";
    }