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Program to find transpose of a matrix

Search element in a sorted matrix

Last Updated : 20 Dec, 2024

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Given a sorted matrix mat[][] of size nxm and an element x, the task is to find if x is present in the matrix or not. Matrix is sorted in a way such that all elements in a row are sorted in increasing order and for row i, where 1 <= i <= n-1, the first element of row i is greater than or equal to the last element of row i-1.

Examples:

Input: x = 14, mat[][] = [[ 1, 5, 9],
[14, 20, 21],
[30, 34, 43]]
Output: true
Input: x = 42, mat[][] = [[ 1, 5, 9, 11],
[14, 20, 21, 26],
[30, 34, 43, 50]]
Output: false

Table of Content

[Naive Approach] Comparing with all elements – O(n*m) Time and O(1) Space
[Better Approach] Using Binary Search Twice - O(logn + logm) Time and O(1) Space
[Expected Approach] Using Binary Search Once - O(log(n*m)) and O(1) Space

**[Naive Approach] Comparing with all elements – O(n*m) Time and O(1) Space**

In this approach, we will iterate through the entire matrix mat[][] and compare each element with x. If an element matches the x, we will return true. Otherwise, at the end of the traversal, we will return false. Please refer to Searching Algorithms for 2D Arrays (Matrix) to know more about the implementation.

[Better Approach] Using Binary Search Twice - O(logn + logm) Time and O(1) Space

First, we locate the row where the target x might be by using binary search, and then we apply binary search again within that row.

To find the correct row, we perform binary search on the first elements of the middle row.
Start with low = 0 and high = n - 1.
If x is smaller than the first element of the middle row (a[mid][0]), then x will be smaller than all elements in rows >= mid, so update high = mid - 1.
If x is greater than the first element of the middle row (a[mid][0]), then x will be greater than all elements in rows < mid, so store the current mid row and update low = mid + 1.
Once we have found the correct row, we can apply binary search within that row to search for the target element x.

C++

// C++ program to search in the sorted matrix using // Binary Search two times  #include <iostream> #include <vector> using namespace std;  // Function to binary search for x in arr[] bool search(vector<int> &arr, int x) {     int lo = 0, hi = arr.size() - 1;      while (lo <= hi) {         int mid = (lo + hi) / 2;          if (x == arr[mid])             return true;         if (x < arr[mid])             hi = mid - 1;         else             lo = mid + 1;     }     return false; }  bool searchMatrix(vector<vector<int>> &mat, int x) {     int n = mat.size(), m = mat[0].size();     int lo = 0, hi = n - 1;     int row = -1;      while (lo <= hi) {         int mid = (lo + hi) / 2;          // If the first element of mid row is equal to x,         // return true         if (x == mat[mid][0])             return true;                // If x is greater than first element of mid row,         // store the mid row and search in lower half         if (x > mat[mid][0]) {             row = mid;             lo = mid + 1;         }                // If x is smaller than first element of mid row,         // search in upper half         else             hi = mid - 1;     }         // If x is smaller than all elements of mat[][]     if (row == -1)         return false;      return search(mat[row], x); }  int main() {     vector<vector<int>> mat = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}};     int x = 14;      if (searchMatrix(mat, x))         cout << "true";     else         cout << "false";      return 0; }

Java

// Java program to search in the sorted matrix using // Binary Search two times  class GfG {          // Function to binary search for x in arr[]     static boolean search(int[] arr, int x) {         int lo = 0, hi = arr.length - 1;          while (lo <= hi) {             int mid = (lo + hi) / 2;              if (x == arr[mid])                 return true;             if (x < arr[mid])                 hi = mid - 1;             else                 lo = mid + 1;         }         return false;     }      static boolean searchMatrix(int[][] mat, int x) {         int n = mat.length, m = mat[0].length;         int lo = 0, hi = n - 1;         int row = -1;          while (lo <= hi) {             int mid = (lo + hi) / 2;              // If the first element of mid row is equal to x,             // return true             if (x == mat[mid][0])                 return true;              // If x is greater than first element of mid row,             // store the mid row and search in lower half             if (x > mat[mid][0]) {                 row = mid;                 lo = mid + 1;             }              // If x is smaller than first element of mid row,             // search in upper half             else                 hi = mid - 1;         }          // If x is smaller than all elements of mat[][]         if (row == -1)             return false;          return search(mat[row], x);     }      public static void main(String[] args) {         int[][] mat = {             {1, 5, 9},             {14, 20, 21},             {30, 34, 43}         };         int x = 14;          if (searchMatrix(mat, x))             System.out.println("true");         else             System.out.println("false");     } }

Python

# Python program to search in the sorted matrix using # Binary Search two times  # Function to binary search for x in arr[] def search(arr, x):     lo = 0     hi = len(arr) - 1      while lo <= hi:         mid = (lo + hi) // 2          if x == arr[mid]:             return True         if x < arr[mid]:             hi = mid - 1         else:             lo = mid + 1     return False  def searchMatrix(mat, x):     n = len(mat)     m = len(mat[0])     lo = 0     hi = n - 1     row = -1      while lo <= hi:         mid = (lo + hi) // 2          # If the first element of mid row is equal to x,         # return true         if x == mat[mid][0]:             return True          # If x is greater than first element of mid row,         # store the mid row and search in lower half         if x > mat[mid][0]:             row = mid             lo = mid + 1          # If x is smaller than first element of mid row,         # search in upper half         else:             hi = mid - 1      # If x is smaller than all elements of mat[][]     if row == -1:         return False      return search(mat[row], x)  if __name__ == "__main__":     mat = [[1, 5, 9], [14, 20, 21], [30, 34, 43]]     x = 14      if searchMatrix(mat, x):         print("true")     else:         print("false")

C#

// C# program to search in the sorted matrix using // Binary Search two times  using System;  class GfG {        // Function to binary search for x in arr[]     static bool Search(int[] arr, int x) {         int lo = 0, hi = arr.Length - 1;          while (lo <= hi) {             int mid = (lo + hi) / 2;              if (x == arr[mid])                 return true;             if (x < arr[mid])                 hi = mid - 1;             else                 lo = mid + 1;         }         return false;     }      static bool SearchMatrix(int[][] mat, int x) {         int n = mat.Length, m = mat[0].Length;         int lo = 0, hi = n - 1;         int row = -1;          while (lo <= hi) {             int mid = (lo + hi) / 2;              // If the first element of mid row is equal to x,             // return true             if (x == mat[mid][0])                 return true;              // If x is greater than first element of mid row,             // store the mid row and search in lower half             if (x > mat[mid][0]) {                 row = mid;                 lo = mid + 1;             }              // If x is smaller than first element of mid row,             // search in upper half             else                 hi = mid - 1;         }          // If x is smaller than all elements of mat[][]         if (row == -1)             return false;          return Search(mat[row], x);     }      static void Main(string[] args) {         int[][] mat = new int[][] {             new int[] {1, 5, 9},             new int[] {14, 20, 21},             new int[] {30, 34, 43}         };         int x = 14;          if (SearchMatrix(mat, x))             Console.WriteLine("true");         else             Console.WriteLine("false");     } }

JavaScript

// JavaScript program to search in the sorted matrix using // Binary Search two times  // Function to binary search for x in arr[] function search(arr, x) {     let lo = 0, hi = arr.length - 1;      while (lo <= hi) {         let mid = Math.floor((lo + hi) / 2);          if (x === arr[mid])             return true;         if (x < arr[mid])             hi = mid - 1;         else             lo = mid + 1;     }     return false; }  function searchMatrix(mat, x) {     let n = mat.length, m = mat[0].length;     let lo = 0, hi = n - 1;     let row = -1;      while (lo <= hi) {         let mid = Math.floor((lo + hi) / 2);          // If the first element of mid row is equal to x,         // return true         if (x === mat[mid][0])             return true;          // If x is greater than first element of mid row,         // store the mid row and search in lower half         if (x > mat[mid][0]) {             row = mid;             lo = mid + 1;         }          // If x is smaller than first element of mid row,         // search in upper half         else             hi = mid - 1;     }      // If x is smaller than all elements of mat[][]     if (row === -1)         return false;      return search(mat[row], x); }  // Driver Code const mat = [     [1, 5, 9],     [14, 20, 21],     [30, 34, 43] ]; const x = 14;  if (searchMatrix(mat, x))     console.log("true"); else     console.log("false");

Output

true

[Expected Approach] Using Binary Search Once - O(log(n*m)) and O(1) Space

The idea is to consider the given matrix as 1D array and apply only one binary search. For example, for a matrix of size n x m and we can consider it as a 1D array of size n*m, then the first index would be 0 and last index would n*m-1. So, we need to do binary search from low = 0 to high = (n*m-1).
How to find the element in 2D matrix corresponding to index = mid?
Since each row of mat[][] will have m elements, so we can find the row of the element as (mid / m) and the column of the element as (mid % m). Then, we can compare x with arr[mid/m][mid%m] for each mid and complete our binary search.

C++

// C++ program to search in the sorted matrix using // binary search  #include <iostream> #include <vector> using namespace std;  bool searchMatrix(vector<vector<int>>& mat, int x) {     int n = mat.size(), m = mat[0].size();      int lo = 0, hi = n * m - 1;     while (lo <= hi) {         int mid = (lo + hi) / 2;                // Find row and column of element at mid index         int row = mid / m;         int col = mid % m;                // If x is found, return true         if (mat[row][col] == x)              return true;                // If x is greater than mat[row][col], search in         // right half         if (mat[row][col] < x)              lo = mid + 1;                  // If x is less than mat[row][col], search in          // left half         else              hi = mid - 1;     }     return false; }  int main() {     vector<vector<int>> mat = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}};     int x = 14;      if (searchMatrix(mat, x))         cout << "true";     else         cout << "false";      return 0; }

Java

// Java program to search in the sorted matrix using // binary search  class GfG {      // Function to search for x in the matrix using binary search     static boolean searchMatrix(int[][] mat, int x) {         int n = mat.length, m = mat[0].length;          int lo = 0, hi = n * m - 1;         while (lo <= hi) {             int mid = (lo + hi) / 2;              // Find row and column of element at mid index             int row = mid / m;             int col = mid % m;              // If x is found, return true             if (mat[row][col] == x)                 return true;              // If x is greater than mat[row][col], search in             // right half             if (mat[row][col] < x)                 lo = mid + 1;              // If x is less than mat[row][col], search in             // left half             else                 hi = mid - 1;         }         return false;     }      public static void main(String[] args) {         int[][] mat = {{1, 5, 9}, {14, 20, 21}, {30, 34, 43}};         int x = 14;          if (searchMatrix(mat, x))             System.out.println("true");         else             System.out.println("false");     } }

Python

# Python program to search in the sorted matrix using # binary search  # Function to search for x in the matrix using binary search def searchMatrix(mat, x):     n = len(mat)     m = len(mat[0])      lo, hi = 0, n * m - 1     while lo <= hi:         mid = (lo + hi) // 2          # Find row and column of element at mid index         row = mid // m         col = mid % m          # If x is found, return true         if mat[row][col] == x:             return True          # If x is greater than mat[row][col], search in         # right half         if mat[row][col] < x:             lo = mid + 1          # If x is less than mat[row][col], search in         # left half         else:             hi = mid - 1      return False  if __name__ == "__main__":     mat = [[1, 5, 9], [14, 20, 21], [30, 34, 43]]     x = 14      if searchMatrix(mat, x):         print("true")     else:         print("false")

C#

// C# program to search in the sorted matrix using // binary search  using System;  class GfG {          // Function to search for x in the matrix using binary search     static bool searchMatrix(int[,] mat, int x) {         int n = mat.GetLength(0), m = mat.GetLength(1);          int lo = 0, hi = n * m - 1;         while (lo <= hi) {             int mid = (lo + hi) / 2;              // Find row and column of element at mid index             int row = mid / m;             int col = mid % m;              // If x is found, return true             if (mat[row, col] == x)                 return true;              // If x is greater than mat[row, col], search in             // right half             if (mat[row, col] < x)                 lo = mid + 1;              // If x is less than mat[row, col], search in             // left half             else                 hi = mid - 1;         }         return false;     }      static void Main() {         int[,] mat = { { 1, 5, 9 }, { 14, 20, 21 }, { 30, 34, 43 } };         int x = 14;          if (searchMatrix(mat, x))             Console.WriteLine("true");         else             Console.WriteLine("false");     } }

JavaScript

// JavaScript program to search in the sorted matrix using // binary search  // Function to search for x in the matrix using binary search function searchMatrix(mat, x) {     let n = mat.length, m = mat[0].length;      let lo = 0, hi = n * m - 1;     while (lo <= hi) {         let mid = Math.floor((lo + hi) / 2);          // Find row and column of element at mid index         let row = Math.floor(mid / m);         let col = mid % m;          // If x is found, return true         if (mat[row][col] === x)             return true;          // If x is greater than mat[row][col], search in         // right half         if (mat[row][col] < x)             lo = mid + 1;          // If x is less than mat[row][col], search in         // left half         else             hi = mid - 1;     }     return false; }  // Driver Code let mat = [[1, 5, 9], [14, 20, 21], [30, 34, 43]]; let x = 14;  if (searchMatrix(mat, x))     console.log("true"); else     console.log("false");

Output

true

Program to find transpose of a matrix

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Article Tags :

Divide and Conquer
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Find number of transformation to make two Matrix Equal

Given two matrices a and b of size n*m. The task is to find the required number of transformation steps so that both matrices become equal. Print -1 if this is not possible. The transformation step is as follows: Select any one matrix out of two matrices. Choose either row/column of the selected mat

Inplace (Fixed space) M x N size matrix transpose

Given an M x N matrix, transpose the matrix without auxiliary memory.It is easy to transpose matrix using an auxiliary array. If the matrix is symmetric in size, we can transpose the matrix inplace by mirroring the 2D array across it's diagonal (try yourself). How to transpose an arbitrary size matr

Minimum flip required to make Binary Matrix symmetric

Given a Binary Matrix mat[][] of size n x n, consisting of 1s and 0s. The task is to find the minimum flips required to make the matrix symmetric along the main diagonal.Examples : Input: mat[][] = [[0, 0, 1], [1, 1, 1], [1, 0, 0]];Output: 2Value of mat[1][0] is not equal to mat[0][1].Value of mat[2

Magic Square of Odd Order

Given a positive integer n, your task is to generate a magic square of order n * n. A magic square of order n is an nâ€¯*â€¯n grid filled with the numbers 1 through nÂ² so that every row, every column, and both main diagonals each add up to the same total, called the magic constant (or magic sum) M. Beca

Hard problems on Matrix

Number of Islands

Given an n x m grid of 'W' (Water) and 'L' (Land), the task is to count the number of islands. An island is a group of adjacent 'L' cells connected horizontally, vertically, or diagonally, and it is surrounded by water or the grid boundary. The goal is to determine how many distinct islands exist in

A Boolean Matrix Question

Given a boolean matrix mat where each cell contains either 0 or 1, the task is to modify it such that if a matrix cell matrix[i][j] is 1 then all the cells in its ith row and jth column will become 1.Examples:Input: [[1, 0], [0, 0]]Output: [[1, 1], [1, 0]]Input: [[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0

Matrix Chain Multiplication

Given the dimension of a sequence of matrices in an array arr[], where the dimension of the ith matrix is (arr[i-1] * arr[i]), the task is to find the most efficient way to multiply these matrices together such that the total number of element multiplications is minimum. When two matrices of size m*

Maximum size rectangle binary sub-matrix with all 1s

Given a 2d binary matrix mat[][], the task is to find the maximum size rectangle binary-sub-matrix with all 1's. Examples: Input: mat = [ [0, 1, 1, 0], [1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 0, 0] ]Output : 8Explanation : The largest rectangle with only 1's is from (1, 0) to (2, 3) which is[1, 1, 1, 1][

Construct Ancestor Matrix from a Given Binary Tree

Given a Binary Tree where all values are from 0 to n-1. Construct an ancestor matrix mat[n][n] where the ancestor matrix is defined as below. mat[i][j] = 1 if i is ancestor of jmat[i][j] = 0, otherwiseExamples: Input: Output: {{0 1 1} {0 0 0} {0 0 0}}Input: Output: {{0 0 0 0 0 0} {1 0 0 0 1 0} {0 0

K'th element in spiral form of matrix

Given a matrix of size n * m. You have to find the kth element which will obtain while traversing the matrix spirally starting from the top-left corner of the matrix.Examples:Input: mat[][] = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ], k = 4Output: 6Explanation: Spiral traversal of matrix: {1, 2, 3, 6, 9,

Largest Plus or '+' formed by all ones in a binary square matrix

Given an n Ã— n binary matrix mat consisting of 0s and 1s. Your task is to find the size of the largest â€˜+â€™ shape that can be formed using only 1s. A â€˜+â€™ shape consists of a center cell with four arms extending in all four directions (up, down, left, and right) while remaining within the matrix bound

Shortest path in a Binary Maze

Given an M x N matrix where each element can either be 0 or 1. We need to find the shortest path between a given source cell to a destination cell. The path can only be created out of a cell if its value is 1.Note: You can move into an adjacent cell in one of the four directions, Up, Down, Left, and

Maximum sum square sub-matrix of given size

Given a 2d array mat[][] of order n * n, and an integer k. Your task is to find a submatrix of order k * k, such that sum of all the elements in the submatrix is maximum possible.Note: Matrix mat[][] contains zero, positive and negative integers.Examples:Input: k = 3mat[][] = [ [ 1, 2, -1, 4 ] [ -8,

Validity of a given Tic-Tac-Toe board configuration

A Tic-Tac-Toe board is given after some moves are played. Find out if the given board is valid, i.e., is it possible to reach this board position after some moves or not.Note that every arbitrary filled grid of 9 spaces isn't valid e.g. a grid filled with 3 X and 6 O isn't valid situation because ea

Minimum Initial Points to Reach Destination

Given a m*n grid with each cell consisting of positive, negative, or no points i.e., zero points. From a cell (i, j) we can move to (i+1, j) or (i, j+1) and we can move to a cell only if we have positive points ( > 0 ) when we move to that cell. Whenever we pass through a cell, points in that cel

Program for Sudoku Generator

Given an integer k, the task is to generate a 9 x 9 Sudoku grid having k empty cells while following the below set of rules:In all 9 submatrices 3x3, the elements should be 1-9, without repetition.In all rows, there should be elements between 1-9, without repetition.In all columns, there should be e

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