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  1. Sherlock and Array
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Sherlock and Array

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  • pradymm
     + 0 comments
    public static String balancedSums(List<Integer> arr) {
// Write your code here
    long total = 0l;
    for(Integer i : arr){
        total = total + i;
    }
    long leftSum =0L;
    long righSum = 0L;
    for(int i= 0; i < arr.size(); i++){
        righSum = total - leftSum - arr.get(i);
        if(righSum == leftSum){
            return "YES";
        }
        leftSum = leftSum + arr.get(i);
    }
    return "NO";
}

  • dalencaste
     + 0 comments

    Java solution using only one loop.

    Time complexity: O(n)

    Space complexity: O(1)

    I found a test case that HackerRank ignores: inputs like "0000...a" should return "YES", but they incorrectly accept "NO" as a valid result.

    public static String balancedSums(List<Integer> arr) {
        int n = arr.size();
        arr.add(0,0);
        arr.add(0);
        int i = 0;
        int j = n+1;
        int cSumL = arr.get(i);
        int cSumR = arr.get(j);
        while(j>=i+2){
            if(cSumL>cSumR){
                j--;
                cSumR += arr.get(j);    
            }else if(cSumL<cSumR){
                i++;
                cSumL += arr.get(i);
            }else{
                if(j==i+2){
                    return "YES";
                }else{
                    if(arr.get(j-1)==0){
                        j--;
                    }else{
                        i++;
                        cSumL += arr.get(i);
                        
                    }
                }
            }

        }
        return "NO";
    }

  • victor_reial
     + 0 comments

    Python best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    def sherlock_and_array(arr):
    # Time complexity: O(n)
    # Space complexity (ignoring input): O(1)
    total_sum = 0
    for value in arr:
        total_sum += value

    left_sum = 0
    for value in arr:
        right_sum = total_sum - left_sum - value
        if left_sum == right_sum:
            return "YES"
        left_sum += value

    return "NO"

  • victor_reial
     + 0 comments

    Rust best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    fn sherlock_and_array(arr: &[i32]) -> String {
//Time complexity: O(n)
    //Space complexity (ignoring input): O(1)
    let total_sum: i32 = arr.iter().sum();

    let mut left_sum = 0;
    for value in arr {
        let right_sum = total_sum - left_sum - value;
        if left_sum == right_sum {
            return "YES".to_string();
        }
        left_sum += value;
    }

    "NO".to_string()

  • shiaramoon
     + 0 comments
    #Python 3
def balancedSums(arr):
    left = 0
    right = sum(arr)
    for val in arr:
        right -=val
        if left == right:
            return "YES"
        left += val
    
    return "NO"