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  1. Sherlock and Array
  2. Discussions

Sherlock and Array

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151 Discussions

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  • philchen2005
     + 1 comment

    how can [2 0 0 0] has expected answer "YES"?

  • dotaikhoa
     + 0 comments

    short c++ solution using prefix sums - i-th element is the sum of all the previous elements + the i-th element of arr.

    string balancedSums(vector<int> arr) {
    bool ok = false;
    
    int n = arr.size();
    vector<int> a(n+1,0);
    
    for (int i = 1; i <= n; i++) {
        a[i] = a[i-1] + arr[i-1];
    }
    
    for (int i = 1; i < n; i++) {
        int l = a[i-1];
        int r = a[n]-a[i];
        ok |= (r==l);
    }
    return ok or n==1 ? "YES" : "NO";
}

  • hank_ring
     + 0 comments
    from functools import reduce

def balancedSums(arr):
    sum_arr = sum(arr)
    return 'YES' if (reduce(lambda a, b: a + (b if a <= sum_arr // 2 else 0), arr) == reduce(lambda a, b: a + (b if a <= sum_arr // 2 else 0), reversed(arr))) else 'NO'

  • brice0103
     + 0 comments

    The test case is wrong! ( 2 0 0 0 & 0 0 2 0)

    Constraints: 1 <= arr[i] <= 2 x 10^4

  • darren80
     + 0 comments

    Javascript

    function balancedSums(arr: number[]): string {
  if (arr.length === 1) {
    return 'YES';
  }
  
  let balanceIdx = Math.floor(arr.length/2);
  let leftSum = arr.slice(0, balanceIdx).reduce((total, val) => total + val, 0);
  let rightSum = arr.slice(balanceIdx+1).reduce((total, val) => total + val, 0);
  
  if (leftSum === rightSum) {
    return 'YES';
  }
  
  const dir = leftSum < rightSum ? 1 : -1;
  
  while (balanceIdx !== -1 && balanceIdx !== arr.length) {
    const prevBalancePt = arr[balanceIdx];
    balanceIdx += dir;
    const newBalancePt = arr[balanceIdx];
    
    if (dir === 1) {
      leftSum += prevBalancePt;
      rightSum -= newBalancePt;
    } else {
      leftSum -= prevBalancePt;
      rightSum += newBalancePt;
    }
    
    if (leftSum === rightSum) {
     return 'YES';
    }
  }
  
  return 'NO';
}