# Kth diagonal from the top left of a given matrix

• Last Updated : 18 Jun, 2021

Given a squared matrix M[ ][ ] of N * N dimensions, the task is to find the Kth diagonal of the matrix, starting from the top-left towards the bottom-right, and print its contents.
Examples:

Input: N = 3, K = 2, M[ ][ ] = {{4, 7, 8}, {9, 2, 3}, {0, 4, 1}
Output: 9 7
Explanation:
5 possible diagonals starting from the top left to the bottom right for the given matrix are:
1 -> {4}
2 -> {9, 7}
3 -> {0, 2, 8}
4 -> {4, 3}
5 -> {1}
Since K = 2, the required diagonal is (9, 7).
Input: N = 2, K = 2, M[ ][ ] = {1, 5}, {9, 4}}
Output:
Explanation:
3 possible diagonals starting from the bottom left to the top right for the given matrix are:
1 -> {1}
2 -> {9, 5}
3 -> {4}
Since K = 2, the required diagonal is (9, 5).

Naive Approach:
The simplest approach to solve this problem is to generate all diagonals starting from the top-left to the bottom right of the given matrix. After generating the Kth diagonal, print its contents.
Time complexity: O(N2
Auxiliary Space: O(N)
Efficient Approach:
The above approach can be optimized by avoiding the traversal of the matrix. Instead, find the boundaries of the Kth diagonal i.e. bottom-left and top-right indices. Once these indices are obtained, print the required diagonal by just traversing the indices of the diagonal.
The following observations are needed to find the position of the diagonal:

If ((K – 1) < N): The diagonal is an upper diagonal
Bottom-left boundary = M[K – 1]
Top-right boundary = M[K – 1]
If (K ? N): The diagonal is a lower diagonal.
Bottom-left boundary = M[N-1][K-N]
Top-right boundary = M[K-N][N-1]

Follow the steps below to solve the problem:

• If (K – 1) < N, set the starting row, startRow as K – 1 and column, startCol as 0.
• Otherwise, set startRow as N – 1 and startCol as K – N.
• Finally, print the elements of diagonal starting from M[startRow][startCol].

Below is the implementation of the above approach:

## C++

 `// C++ implementation of``// the above approach``#include ``using` `namespace` `std;` `// Function returns required diagonal``void` `printDiagonal(``int` `K, ``int` `N,``                ``vector >& M)``{``    ``int` `startrow, startcol;` `    ``// Initialize values to``    ``// print upper diagonals``    ``if` `(K - 1 < N) {``        ``startrow = K - 1;``        ``startcol = 0;``    ``}` `    ``// Initialize values to``    ``// print lower diagonals``    ``else` `{``        ``startrow = N - 1;``        ``startcol = K - N;``    ``}` `    ``// Traverse the diagonal``    ``for` `(; startrow >= 0 && startcol < N;``        ``startrow--, startcol++) {` `        ``// Print its contents``        ``cout << M[startrow][startcol]``            ``<< ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `N = 3, K = 4;``    ``vector > M = { { 4, 7, 8 },``                            ``{ 9, 2, 3 },``                            ``{ 0, 4, 1 } };` `    ``printDiagonal(K, N, M);``    ``return` `0;``}`

## Java

 `// Java implementation of``// the above approach``import` `java.util.*;` `class` `GFG{` `// Function returns required diagonal``static` `void` `printDiagonal(``int` `K, ``int` `N,``                          ``int` `[][]M)``{``    ``int` `startrow, startcol;` `    ``// Initialize values to``    ``// print upper diagonals``    ``if` `(K - ``1` `< N)``    ``{``        ``startrow = K - ``1``;``        ``startcol = ``0``;``    ``}` `    ``// Initialize values to``    ``// print lower diagonals``    ``else``    ``{``        ``startrow = N - ``1``;``        ``startcol = K - N;``    ``}` `    ``// Traverse the diagonal``    ``for``(; startrow >= ``0` `&& startcol < N;``          ``startrow--, startcol++)``    ``{``        ` `        ``// Print its contents``        ``System.out.print(M[startrow][startcol] + ``" "``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``3``, K = ``4``;``    ``int``[][] M = { { ``4``, ``7``, ``8` `},``                  ``{ ``9``, ``2``, ``3` `},``                  ``{ ``0``, ``4``, ``1` `} };` `    ``printDiagonal(K, N, M);``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the``# above approach``def` `printDiagonal(K, N, M):``    ` `    ``startrow, startcol ``=` `0``, ``0` `    ``# Initialize values to prupper``    ``# diagonals``    ``if` `K ``-` `1` `< N:``        ``startrow ``=` `K ``-` `1``        ``startcol ``=` `0``    ``else``:``        ``startrow ``=` `N ``-` `1``        ``startcol ``=` `K ``-` `N` `    ``# Traverse the diagonals``    ``while` `startrow >``=` `0` `and` `startcol < N:``        ` `        ``# Print its contents``        ``print``(M[startrow][startcol], end ``=` `" "``)``        ``startrow ``-``=` `1``        ``startcol ``+``=` `1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``N, K ``=` `3``, ``4``    ``M ``=` `[ [ ``4``, ``7``, ``8` `],``          ``[ ``9``, ``2``, ``3` `],``          ``[ ``0``, ``4``, ``1` `] ]``          ` `    ``printDiagonal(K, N, M)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of``// the above approach``using` `System;``class` `GFG{` `// Function returns required diagonal``static` `void` `printDiagonal(``int` `K, ``int` `N,``                          ``int` `[,]M)``{``    ``int` `startrow, startcol;` `    ``// Initialize values to``    ``// print upper diagonals``    ``if` `(K - 1 < N)``    ``{``        ``startrow = K - 1;``        ``startcol = 0;``    ``}` `    ``// Initialize values to``    ``// print lower diagonals``    ``else``    ``{``        ``startrow = N - 1;``        ``startcol = K - N;``    ``}` `    ``// Traverse the diagonal``    ``for``(; startrow >= 0 && startcol < N;``          ``startrow--, startcol++)``    ``{``        ` `        ``// Print its contents``        ``Console.Write(M[startrow,startcol] + ``" "``);``    ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 3, K = 4;``    ``int``[,] M = { { 4, 7, 8 },``                 ``{ 9, 2, 3 },``                 ``{ 0, 4, 1 } };` `    ``printDiagonal(K, N, M);``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`4 3 `

Time complexity: O(N)
Auxiliary Space: O(1)

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