Print matrix in diagonal pattern - GeeksforGeeks

07 Feb, 2022
Given a matrix of n*n size, the task is to print its elements in a diagonal pattern.  
 
Input : mat[3][3] = {{1, 2, 3},
                     {4, 5, 6},
                     {7, 8, 9}}
Output : 1 2 4 7 5 3 6 8 9.
Explanation: Start from 1 
Then from upward to downward diagonally i.e. 2 and 4
Then from downward to upward diagonally i.e 7, 5, 3 
Then from up to down diagonally i.e  6, 8 
Then down to up i.e. end at 9.

Input :  mat[4][4] =  {{1,  2,  3,  10},
                      {4,  5,  6,  11},
                      {7,  8,  9,  12},
                      {13, 14, 15, 16}}
Output:  1 2 4 7 5 3 10 6 8 13 14 9 11 12 15 16 .
Explanation: Start from 1 
Then from upward to downward diagonally i.e. 2 and 4
Then from downward to upward diagonally i.e 7, 5, 3 
Then from upward to downward diagonally i.e. 10 6 8 13
Then from downward to upward diagonally i.e 14 9 11
Then from upward to downward diagonally i.e. 12 15
then end at 16
 
Approach: From the diagram it can be seen that every element is either printed diagonally upward or diagonally downward. Start from the index (0,0) and print the elements diagonally upward then change the direction, change the column and print diagonally downwards. This cycle continues until the last element is reached.Algorithm:  
Create variables i=0, j=0 to store the current indices of row and columnRun a loop from 0 to n*n, where n is side of the matrix.Use a flag isUp to decide whether the direction is upwards or downwards. Set isUp = true initially the direction is going upward.If isUp = 1 then start printing elements by incrementing column index and decrementing the row index.Similarly if isUp = 0, then decrement the column index and increment the row index.Move to the next column or row (next starting row and columnDo this till all the elements get traversed.
Create variables i=0, j=0 to store the current indices of row and column
Run a loop from 0 to n*n, where n is side of the matrix.
Use a flag isUp to decide whether the direction is upwards or downwards. Set isUp = true initially the direction is going upward.
If isUp = 1 then start printing elements by incrementing column index and decrementing the row index.
Similarly if isUp = 0, then decrement the column index and increment the row index.
Move to the next column or row (next starting row and column
Do this till all the elements get traversed.
Implementation:  
C++
Java
Python 3
C#
PHP
Javascript
// C++ program to print matrix in diagonal order#include <bits/stdc++.h>using namespace std;const int MAX = 100; void printMatrixDiagonal(int mat[MAX][MAX], int n){    // Initialize indexes of element to be printed next    int i = 0, j = 0;     // Direction is initially from down to up    bool isUp = true;     // Traverse the matrix till all elements get traversed    for (int k = 0; k < n * n;) {        // If isUp = true then traverse from downward        // to upward        if (isUp) {            for (; i >= 0 && j < n; j++, i--) {                cout << mat[i][j] << " ";                k++;            }             // Set i and j according to direction            if (i < 0 && j <= n - 1)                i = 0;            if (j == n)                i = i + 2, j--;        }         // If isUp = 0 then traverse up to down        else {            for (; j >= 0 && i < n; i++, j--) {                cout << mat[i][j] << " ";                k++;            }             // Set i and j according to direction            if (j < 0 && i <= n - 1)                j = 0;            if (i == n)                j = j + 2, i--;        }         // Revert the isUp to change the direction        isUp = !isUp;    }} int main(){    int mat[MAX][MAX] = { { 1, 2, 3 },                          { 4, 5, 6 },                          { 7, 8, 9 } };     int n = 3;    printMatrixDiagonal(mat, n);    return 0;}
// Java program to print matrix in diagonal orderclass GFG {    static final int MAX = 100;     static void printMatrixDiagonal(int mat[][], int n)    {        // Initialize indexes of element to be printed next        int i = 0, j = 0;         // Direction is initially from down to up        boolean isUp = true;         // Traverse the matrix till all elements get traversed        for (int k = 0; k < n * n;) {            // If isUp = true then traverse from downward            // to upward            if (isUp) {                for (; i >= 0 && j < n; j++, i--) {                    System.out.print(mat[i][j] + " ");                    k++;                }                 // Set i and j according to direction                if (i < 0 && j <= n - 1)                    i = 0;                if (j == n) {                    i = i + 2;                    j--;                }            }             // If isUp = 0 then traverse up to down            else {                for (; j >= 0 && i < n; i++, j--) {                    System.out.print(mat[i][j] + " ");                    k++;                }                 // Set i and j according to direction                if (j < 0 && i <= n - 1)                    j = 0;                if (i == n) {                    j = j + 2;                    i--;                }            }             // Revert the isUp to change the direction            isUp = !isUp;        }    }     // Driver code    public static void main(String[] args)    {        int mat[][] = { { 1, 2, 3 },                        { 4, 5, 6 },                        { 7, 8, 9 } };         int n = 3;        printMatrixDiagonal(mat, n);    }}// This code is contributed by Anant Agarwal.
# Python 3 program to print matrix in diagonal orderMAX = 100 def printMatrixDiagonal(mat, n):    # Initialize indexes of element to be printed next    i = 0    j = 0    k = 0    # Direction is initially from down to up    isUp = True      # Traverse the matrix till all elements get traversed    while k<n * n:         # If isUp = True then traverse from downward         # to upward        if isUp:            while i >= 0 and j<n :                print(str(mat[i][j]), end = " ")                k += 1                j += 1                i -= 1               # Set i and j according to direction            if i < 0 and j <= n - 1:                i = 0            if j == n:                i = i + 2                j -= 1          # If isUp = 0 then traverse up to down        else:            while j >= 0 and i<n :                print(mat[i][j], end = " ")                k += 1                i += 1                j -= 1               # Set i and j according to direction            if j < 0 and i <= n - 1:                j = 0            if i == n:                j = j + 2                i -= 1          # Revert the isUp to change the direction        isUp = not isUp # Driver programif __name__ == "__main__":    mat = [[1, 2, 3],        [4, 5, 6],        [7, 8, 9] ]    n = 3   printMatrixDiagonal(mat, n) # This code is contributed by Chitra Nayal
// C# program to print matrix in diagonal orderusing System;class GFG {    static int MAX = 100;     static void printMatrixDiagonal(int[, ] mat, int n)    {        // Initialize indexes of element to be printed next        int i = 0, j = 0;         // Direction is initially from down to up        bool isUp = true;         // Traverse the matrix till all elements get traversed        for (int k = 0; k < n * n;) {            // If isUp = true then traverse from downward            // to upward            if (isUp) {                for (; i >= 0 && j < n; j++, i--) {                    Console.Write(mat[i, j] + " ");                    k++;                }                 // Set i and j according to direction                if (i < 0 && j <= n - 1)                    i = 0;                if (j == n) {                    i = i + 2;                    j--;                }            }             // If isUp = 0 then traverse up to down            else {                for (; j >= 0 && i < n; i++, j--) {                    Console.Write(mat[i, j] + " ");                    k++;                }                 // Set i and j according to direction                if (j < 0 && i <= n - 1)                    j = 0;                if (i == n) {                    j = j + 2;                    i--;                }            }             // Revert the isUp to change the direction            isUp = !isUp;        }    }     // Driver code    public static void Main()    {        int[, ] mat = { { 1, 2, 3 },                        { 4, 5, 6 },                        { 7, 8, 9 } };         int n = 3;        printMatrixDiagonal(mat, n);    }}// This code is contributed by vt_m.
<?php// php program to print matrix// in diagonal order   $MAX = 100;   function printMatrixDiagonal($mat, $n){           // Initialize indexes of element    // to be printed next    $i = 0;    $j = 0 ;       // Direction is initially    // from down to up    $isUp = true;       // Traverse the matrix till    // all elements get traversed    for ($k = 0;$k < $n * $n    {        // If isUp = true then traverse         // from downward to upward        if ($isUp)        {            for ( ;$i >= 0 && $j < $n;$j++, $i--)            {                echo $mat[$i][$j]." ";                $k++;            }               // Set i and j according             // to direction            if ($i < 0 && $j <= $n - 1)                $i = 0;            if ($j == $n)            {                $i = $i + 2;                $j--;            }        }           // If isUp = 0 then         // traverse up to down        else        {            for ( ; $j >= 0 &&                  $i<$n ; $i++, $j--)            {                echo $mat[$i][$j]." ";                $k++;            }               // Set i and j according            // to direction            if ($j < 0 && $i <= $n - 1)                $j = 0;            if ($i == $n)            {                $j = $j + 2;                $i--;            }        }           // Revert the isUp to        // change the direction        $isUp = !$isUp;    }}       // Driver code    $mat= array(array(1, 2, 3),          array(4, 5, 6),          array(7, 8, 9));       $n = 3;    printMatrixDiagonal($mat, $n);   // This code is contributed by Surbhi Tyagi ?>
<script> // Javascript program to print// matrix in diagonal order function printMatrixDiagonal(arr, len) {   // Initialize indices to traverse  // through the array  let i = 0, j = 0;   // Direction is initially from  // down to up  let isUp = true;   // Traverse the matrix till all  // elements get traversed  for (let k = 0; k < len * len;) {     // If isUp = true traverse from    // bottom to top      if (isUp) {          for (;i >= 0 && j < len; i--, j++) {              document.write(arr[i][j] + ' ');              k++;          }           // Set i and j according to direction          if (i < 0 && j < len) i = 0;          if (j === len) i = i + 2, j--;      }       // If isUp = false then traverse      // from top to bottom      else {          for (;j >= 0 && i < len; i++, j--) {            document.write(arr[i][j] + ' ');              k++;          }             // Set i and j according to direction            if (j < 0 && i < len) j = 0;          if (i === len) j = j + 2, i--;      }             // Inverse the value of isUp to      // change the direction      isUp = !isUp  } } // function calllet arr = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];let arrLength = arr.length; printMatrixDiagonal(arr, arrLength); // This code is contributed by karthiksrinivasprasad </script>
1 2 4 7 5 3 6 8 9
 
Complexity Analysis:  
Time Complexity: O(n*n). To traverse the matrix O(n*n) time complexity is needed.
Space Complexity: O(1). As no extra space is required.
Alternate Implementation: This is another simple and compact implementation of the same approach as mentioned above. 
C++
Java
Python3
C#
Javascript
// C++ program to print matrix in diagonal order#include <bits/stdc++.h>using namespace std; int main(){    // Initialize matrix    int mat[][4] = { { 1, 2, 3, 4 },                     { 5, 6, 7, 8 },                     { 9, 10, 11, 12 },                     { 13, 14, 15, 16 } };    // n - size    // mode - switch to derive up/down traversal    // it - iterator count - increases until it    // reaches n and then decreases    int n = 4, mode = 0, it = 0, lower = 0;     // 2n will be the number of iterations    for (int t = 0; t < (2 * n - 1); t++) {        int t1 = t;        if (t1 >= n) {            mode++;            t1 = n - 1;            it--;            lower++;        }        else {            lower = 0;            it++;        }        for (int i = t1; i >= lower; i--) {            if ((t1 + mode) % 2 == 0) {                cout << (mat[i][t1 + lower - i]) << endl;            }            else {                cout << (mat[t1 + lower - i][i]) << endl;            }        }    }    return 0;} // This code is contributed by princiraj1992
// Java program to print matrix in diagonal orderpublic class MatrixDiag {     public static void main(String[] args)    {        // Initialize matrix        int[][] mat = { { 1, 2, 3, 4 },                        { 5, 6, 7, 8 },                        { 9, 10, 11, 12 },                        { 13, 14, 15, 16 } };        // n - size        // mode - switch to derive up/down traversal        // it - iterator count - increases until it        // reaches n and then decreases        int n = 4, mode = 0, it = 0, lower = 0;         // 2n will be the number of iterations        for (int t = 0; t < (2 * n - 1); t++) {            int t1 = t;            if (t1 >= n) {                mode++;                t1 = n - 1;                it--;                lower++;            }            else {                lower = 0;                it++;            }            for (int i = t1; i >= lower; i--) {                if ((t1 + mode) % 2 == 0) {                    System.out.println(mat[i][t1 + lower - i]);                }                else {                    System.out.println(mat[t1 + lower - i][i]);                }            }        }    }}
# Python3 program to print matrix in diagonal order # Driver code # Initialize matrixmat = [[ 1, 2, 3, 4 ],       [ 5, 6, 7, 8 ],       [ 9, 10, 11, 12 ],       [ 13, 14, 15, 16 ]];        # n - size# mode - switch to derive up/down traversal# it - iterator count - increases until it# reaches n and then decreasesn = 4mode = 0it = 0lower = 0 # 2n will be the number of iterationsfor t in range(2 * n - 1):    t1 = t    if (t1 >= n):        mode += 1        t1 = n - 1        it -= 1        lower += 1    else:        lower = 0        it += 1     for i in range(t1, lower - 1, -1):        if ((t1 + mode) % 2 == 0):            print((mat[i][t1 + lower - i]))        else:            print(mat[t1 + lower - i][i]) # This code is contributed by princiraj1992
// C# program to print matrix in diagonal orderusing System; public class MatrixDiag {    // Driver code    public static void Main(String[] args)    {        // Initialize matrix        int[, ] mat = { { 1, 2, 3, 4 },                        { 5, 6, 7, 8 },                        { 9, 10, 11, 12 },                        { 13, 14, 15, 16 } };        // n - size        // mode - switch to derive up/down traversal        // it - iterator count - increases until it        // reaches n and then decreases        int n = 4, mode = 0, it = 0, lower = 0;         // 2n will be the number of iterations        for (int t = 0; t < (2 * n - 1); t++) {            int t1 = t;            if (t1 >= n) {                mode++;                t1 = n - 1;                it--;                lower++;            }            else {                lower = 0;                it++;            }            for (int i = t1; i >= lower; i--) {                if ((t1 + mode) % 2 == 0) {                    Console.WriteLine(mat[i, t1 + lower - i]);                }                else {                    Console.WriteLine(mat[t1 + lower - i, i]);                }            }        }    }} // This code contributed by Rajput-Ji
<script>// Javascript program to print an n * n matrix in diagonal order function MatrixDiag(arr){    // n - size of the array    // mode - to switch from top/bottom traversal    // it - iterator count - increases until it    // reaches n and then decreases    // lower - to ensure we move to the    // next row when columns go out of bounds    let n = arr.length, mode = 0, it = 0, lower = 0;       // A 4 * 4 matrix has 7 diagonals.    // Hence (2 * n -1) iterations       for(let t=0; t< (2 * n - 1); t++){      let t1 = t;      if(t1 >= n){        mode++;        t1 = n - 1;        it--;        lower++;             } else {        lower = 0;        it++;      }      for(let i = t1; i>= lower; i--){        if((t1 + mode) % 2 === 0){          console.log(arr[i][t1 + lower - i])        } else{          console.log(arr[t1 + lower - i][i])        }      }    }   }   // function call  let arr = [                  [1, 2, 3, 4],                [5, 6, 7, 8],                [9, 10, 11, 12],                [13, 14, 15, 16]            ];  MatrixDiag(arr)   // This code is contributed by karthiksrinivasprasad </script>
1
2
5
9
6
3
4
7
10
13
14
11
8
12
15
16
 
This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Mithun Kumar
ukasp
SambaSivaKrishna
Rajput-Ji
princiraj1992
mohit kumar 29
Code_r
andrew1234
surbhityagi15
karthiksrinivasprasad
saurabh1990aror
simmytarika5
surinderdawra388
Amazon
pattern-printing
Matrix
School Programming
Amazon
pattern-printing
Matrix
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Sudoku | Backtracking-7
Divide and Conquer | Set 5 (Strassen's Matrix Multiplication)
Program to multiply two matrices
Inplace rotate square matrix by 90 degrees | Set 1
Min Cost Path | DP-6
Python Dictionary
Arrays in C/C++
Inheritance in C++
Reverse a string in Java
C++ Classes and Objects