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  1. Sherlock and Array
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Sherlock and Array

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  • gewinum
     + 0 comments

    https://pastebin.com/A7Vv1RUK really simple solution made in C#

  • daniel_alvarado2
     + 0 comments

    C#

    public static string balancedSums(List arr) {

        int sumLeft = 0;
    int sumRight = arr.Skip(1).Sum();

    for(int i = 0; i < arr.Count; i++)
    {  
        if(i>0)
        {          
            sumLeft += arr[i - 1];
            sumRight -= arr[i];
        }             

        if(sumLeft == sumRight)
        {
            return "YES";
        }     
    }

    return "NO";
}

  • ApolionXD
     + 0 comments

    Java:

    I don't know what I'm doing wrong:

    public static String balancedSums(List<Integer> arr) {
        Integer sumRight = arr.get(0);
        Integer sumLeft = 0;
        if(arr.size()>2){
            for(int j=2;j<arr.size();j++){
                sumLeft += arr.get(j);
            }
            if(sumLeft==0)
                return "YES";
            for(int i=1;i<(arr.size()-1);i++){
                if(sumRight==sumLeft){
                    return "YES";
                }
                sumLeft-=arr.get(i+1);
                sumRight+=arr.get(i);
            }
        } else if (arr.size()==1){
            return "YES";
        }

        return "NO";
    }

  • emailthegyro
     + 1 comment

    C++

    string balancedSums(vector<int> arr) {
    
    int sums = accumulate(arr.begin(),arr.end(),0);
    int leftSums = 0;
    
    for(int i : arr){
        if(sums-i == leftSums)
            return "YES";
        else{
            sums -= i;
            leftSums += i;
        }
    }
    
    return "NO";
}

  • emailthegyro
     + 0 comments

    Python solution

    def balancedSums(arr):
    sums = sum(arr)
    leftSum = 0
    for i in arr:
        if sums-i == leftSum:
            return "YES"   
        else:
            sums -= i
            leftSum += i   
    return "NO"