We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
Loading...
  1. Practice
  2. Algorithms
  3. Warmup
  4. Diagonal Difference
  5. Discussions

Diagonal Difference

  • misgevolution + 5 comments
    int sum1=0,sum2=0;   

for(int i=0, k=size-1; i< size; i++){
    for(int j=0; j< size; j++){
        cin >> matrix[i][j];            
        if(i==j)
         sum1 += matrix[i][j];                        
    }
    sum2 += matrix[i][k];              
    k -= 1;
}

cout << abs(sum1-sum2);
    • Joxebus + 0 comments
      [deleted]
    • souman1729 + 0 comments

      good thinking

    • jefflima + 2 comments

      Really nice one.

      You can actually move up the k -= 1 into the outer loop next to the i++ as:

      int sum1=0,sum2=0;   

for(int i=0, k=size-1; i< size; i++, k--){
    for(int j=0; j< size; j++){
        cin >> matrix[i][j];            
        if(i==j)
         sum1 += matrix[i][j];                        
    }
    sum2 += matrix[i][k];              
}

cout << abs(sum1-sum2);
      • misgevolution + 0 comments

        Yes, thanks!

      • Deeps79 + 2 comments

        what's the use of k here?? k=size-1...and k-- ? can Anyone help out..please

        • abhijith4514 + 0 comments

          Try writing a matrix in a paper and also give values to i,j,k as given above. You will understand why we need "k" much better that way, than just reading the code.

        • aviral16austy + 2 comments

          yes I dont get k? i reprents rows,j represents columns i think but k?

          • aman1996esseffc1 + 0 comments

            well you can use i in place of k that's not a problem that's just a variable used for looping

          • Kanahaiya + 0 comments

            Here is the video explaination- https://youtu.be/f6bTIsj9ne8

    • vinay666 + 0 comments

      good

    • lemueluhuru + 3 comments

      Nice. You can remove the conditional though, and complete this in one loop. Here is my solution in JavaScript. 

      let primaryDiag = 0
let secondDiag = 0
let length = Matrix.length

for (var i = 0, indexLength = length - 1; i < length; i++) {
    primaryDiag += Matrix[i][i]
    secondaryDiag += Matrix[i][indexLength]
    indexLength -= 1 
}

return Math.abs(primaryDiag - secondaryDiag)
      • tinla14 + 0 comments

        Hello lemuel,

        Can you please explain more to me about this????

      • jonathan_m_whit1 + 0 comments

        I altered a little. 

        function diagonalDifference(arr) {  
        let primDiag = 0
        let secDiag = 0
        let n = arr.length

        for (var i = 0, j = n - 1; i < n; i++, j--) {
                primDiag += arr[i][i]
                secDiag += arr[i][j]  
        }
        return Math.abs(primDiag - secDiag)

        }

      • suhailmdcse + 0 comments

        super!! logic