Given a matrix of m*n size, the task is to count all the rows in a matrix that are sorted either in strictly increasing order or in strictly decreasing order?
Examples :
Input : m = 4, n = 5
mat[m][n] = 1 2 3 4 5
4 3 1 2 6
8 7 6 5 4
5 7 8 9 10
Output: 3
The idea is simple and involves two traversals of matrix.
- Traverse from left side of the matrix to count all the row which are in strictly increasing order
- Traverse from right side of the matrix to count all the row which are in strictly decreasing order
Below is the implementation of above idea.
C++
// C++ program to find number of sorted rows#include <bits/stdc++.h>#define MAX 100using namespace std;// Function to count all sorted rows in a matrixint sortedCount(int mat[][MAX], int r, int c){ int result = 0; // Initialize result // Start from left side of matrix to // count increasing order rows for (int i=0; i<r; i++) { // Check if there is any pair of element // that are not in increasing order. int j; for (j=0; j<c-1; j++) if (mat[i][j+1] <= mat[i][j]) break; // If the loop didn't break (All elements // of current row were in increasing order) if (j == c-1) result++; } // Start from right side of matrix to // count increasing order rows ( reference // to left these are in decreasing order ) for (int i=0; i<r; i++) { // Check if there is any pair of element // that are not in decreasing order. int j; for (j=c-1; j>0; j--) if (mat[i][j-1] <= mat[i][j]) break; // Note c > 1 condition is required to make // sure that a single column row is not counted // twice (Note that a single column row is sorted // both in increasing and decreasing order) if (c > 1 && j == 0) result++; } return result;}// Driver program to run the caseint main(){ int m = 4, n = 5; int mat[][MAX] = {{1, 2, 3, 4, 5}, {4, 3, 1, 2, 6}, {8, 7, 6, 5, 4}, {5, 7, 8, 9, 10}}; cout << sortedCount(mat, m, n); return 0;} |
Java
// Java program to find number of sorted rowsclass GFG { static int MAX = 100; // Function to count all sorted rows in a matrix static int sortedCount(int mat[][], int r, int c) { int result = 0; // Initialize result // Start from left side of matrix to // count increasing order rows for (int i = 0; i < r; i++) { // Check if there is any pair of element // that are not in increasing order. int j; for (j = 0; j < c - 1; j++) if (mat[i][j + 1] <= mat[i][j]) break; // If the loop didn't break (All elements // of current row were in increasing order) if (j == c - 1) result++; } // Start from right side of matrix to // count increasing order rows ( reference // to left these are in decreasing order ) for (int i = 0; i < r; i++) { // Check if there is any pair of element // that are not in decreasing order. int j; for (j = c - 1; j > 0; j--) if (mat[i][j - 1] <= mat[i][j]) break; // Note c > 1 condition is required to make // sure that a single column row is not counted // twice (Note that a single column row is sorted // both in increasing and decreasing order) if (c > 1 && j == 0) result++; } return result; } // Driver code public static void main(String arg[]) { int m = 4, n = 5; int mat[][] = { { 1, 2, 3, 4, 5 }, { 4, 3, 1, 2, 6 }, { 8, 7, 6, 5, 4 }, { 5, 7, 8, 9, 10 } }; System.out.print(sortedCount(mat, m, n)); }}// This code is contributed by Anant Agarwal. |
Python
# Python3 program to find number # of sorted rowsdef sortedCount(mat, r, c): result = 0 # Start from left side of matrix to # count increasing order rows for i in range(r): # Check if there is any pair of element # that are not in increasing order. j = 0 for j in range(c - 1): if mat[i][j + 1] <= mat[i][j]: break # If the loop didn't break (All elements # of current row were in increasing order) if j == c - 2: result += 1 # Start from right side of matrix to # count increasing order rows ( reference # to left these are in decreasing order ) for i in range(0, r): # Check if there is any pair of element # that are not in decreasing order. j = 0 for j in range(c - 1, 0, -1): if mat[i][j - 1] <= mat[i][j]: break # Note c > 1 condition is required to # make sure that a single column row # is not counted twice (Note that a # single column row is sorted both # in increasing and decreasing order) if c > 1 and j == 1: result += 1 return result # Driver codem, n = 4, 5mat = [[1, 2, 3, 4, 5], [4, 3, 1, 2, 6], [8, 7, 6, 5, 4], [5, 7, 8, 9, 10]]print(sortedCount(mat, m, n))# This code is contributed by# Mohit kumar 29 (IIIT gwalior) |
C#
// C# program to find number of sorted rowsusing System;class GFG { // static int MAX = 100; // Function to count all sorted rows in // a matrix static int sortedCount(int [,]mat, int r, int c) { int result = 0; // Initialize result // Start from left side of matrix to // count increasing order rows for (int i = 0; i < r; i++) { // Check if there is any pair of // element that are not in // increasing order. int j; for (j = 0; j < c - 1; j++) if (mat[i,j + 1] <= mat[i,j]) break; // If the loop didn't break (All // elements of current row were // in increasing order) if (j == c - 1) result++; } // Start from right side of matrix // to count increasing order rows // ( reference to left these are in // decreasing order ) for (int i = 0; i < r; i++) { // Check if there is any pair // of element that are not in // decreasing order. int j; for (j = c - 1; j > 0; j--) if (mat[i,j - 1] <= mat[i,j]) break; // Note c > 1 condition is // required to make sure that a // single column row is not // counted twice (Note that a // single column row is sorted // both in increasing and // decreasing order) if (c > 1 && j == 0) result++; } return result; } // Driver code public static void Main() { int m = 4, n = 5; int [,]mat = { { 1, 2, 3, 4, 5 }, { 4, 3, 1, 2, 6 }, { 8, 7, 6, 5, 4 }, { 5, 7, 8, 9, 10 } }; Console.WriteLine( sortedCount(mat, m, n)); }}// This code is contributed by anuj_67. |
PHP
<?php// PHP program to find // number of sorted rows$MAX = 100;// Function to count all // sorted rows in a matrixfunction sortedCount($mat, $r, $c){ // Initialize result $result = 0; // Start from left side of // matrix to count increasing // order rows for ( $i = 0; $i < $r; $i++) { // Check if there is any // pair of element that // are not in increasing order. $j; for ($j = 0; $j < $c - 1; $j++) if ($mat[$i][$j + 1] <= $mat[$i][$j]) break; // If the loop didn't break // (All elements of current // row were in increasing order) if ($j == $c - 1) $result++; } // Start from right side of // matrix to count increasing // order rows ( reference to left // these are in decreasing order ) for ($i = 0; $i < $r; $i++) { // Check if there is any pair // of element that are not // in decreasing order. $j; for ($j = $c - 1; $j > 0; $j--) if ($mat[$i][$j - 1] <= $mat[$i][$j]) break; // Note c > 1 condition is // required to make sure that // a single column row is not // counted twice (Note that a // single column row is sorted // both in increasing and // decreasing order) if ($c > 1 && $j == 0) $result++; } return $result;}// Driver Code$m = 4; $n = 5;$mat = array(array(1, 2, 3, 4, 5), array(4, 3, 1, 2, 6), array(8, 7, 6, 5, 4), array(5, 7, 8, 9, 10));echo sortedCount($mat, $m, $n);// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript program to find number of sorted rows let MAX = 100; // Function to count all sorted rows in a matrix function sortedCount(mat,r,c) { let result = 0; // Initialize result // Start from left side of matrix to // count increasing order rows for (let i = 0; i < r; i++) { // Check if there is any pair of element // that are not in increasing order. let j; for (j = 0; j < c - 1; j++) if (mat[i][j + 1] <= mat[i][j]) break; // If the loop didn't break (All elements // of current row were in increasing order) if (j == c - 1) result++; } // Start from right side of matrix to // count increasing order rows ( reference // to left these are in decreasing order ) for (let i = 0; i < r; i++) { // Check if there is any pair of element // that are not in decreasing order. let j; for (j = c - 1; j > 0; j--) if (mat[i][j - 1] <= mat[i][j]) break; // Note c > 1 condition is required to make // sure that a single column row is not counted // twice (Note that a single column row is sorted // both in increasing and decreasing order) if (c > 1 && j == 0) result++; } return result; } // Driver code let m = 4, n = 5; let mat = [[1, 2, 3, 4, 5], [4, 3, 1, 2, 6], [8, 7, 6, 5, 4], [5, 7, 8, 9, 10]] document.write(sortedCount(mat, m, n)) // This code is contributed by unknown2108</script> |
Output
3
Time Complexity : O(m*n)
Auxiliary space : O(1)
If you have another optimized approach to solve this problem then please share in comments.
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