Search in a row wise and column wise sorted matrix (original) (raw)

Last Updated : 23 Jul, 2025

Given a matrix mat[][] and an integer **x, the task is to check if **x is present in **mat[][] or not. Every row and column of the matrix is sorted in increasing order.

**Examples:

**Input: x = 62, mat[][] = [[3, 30, 38],
[20, 52, 54],
[35, 60, 69]]
**Output: false
**Explanation: 62 is not present in the matrix.

**Input: x = 55, mat[][] = [[18, 21, 27],
[38, 55, 67]]
**Output: true
**Explanation: mat[1][1] is equal to 55.

**Input: x = 35, mat[][] = [[3, 30, 38],
[20, 52, 54],
[35, 60, 69]]
**Output: true
**Explanation: mat[2][0] is equal to 35.

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Table of Content

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

The simple idea is to traverse the complete matrix and search for the target element. If the target element is found, return true. Otherwise, return false.

C++ `

// C++ program to search an element in row-wise // and column-wise sorted matrix

#include #include

using namespace std;

bool matSearch(vector<vector> &mat, int x) { int n = mat.size(), m = mat[0].size();

// Iterate over all the elements to find x
for(int i = 0; i < n; i++) {
    for(int j = 0; j < m; j++) {
        if(mat[i][j] == x)
            return true;
    }
}

// If x was not found, return false
return false;

}

int main() { vector<vector> mat = {{3, 30, 38}, {20, 52, 54}, {35, 60, 69}}; int x = 35; if(matSearch(mat, x)) cout << "true"; else cout << "false"; return 0; }

Java

// Java program to search an element in row-wise // and column-wise sorted matrix

class GfG { static boolean matSearch(int[][] mat, int x) { int n = mat.length, m = mat[0].length;

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(mat[i][j] == x)
                return true;
        }
    }
  
    // If x was not found, return false
    return false;
}

public static void main(String[] args) {
    int[][] mat = {{3, 30, 38},
                   {20, 52, 54},
                   {35, 60, 69}};
    int x = 35;
    
    if(matSearch(mat, x)) 
        System.out.println("true");
    else 
        System.out.println("false");
}

}

Python

Python program to search an element in row-wise

and column-wise sorted matrix

def matSearch(mat, x): n = len(mat) m = len(mat[0])

for i in range(n):
    for j in range(m):
        if mat[i][j] == x:
            return True

# If x was not found, return false
return False

if name == "main": mat = [[3, 30, 38], [20, 52, 54], [35, 60, 69]] x = 35 if matSearch(mat, x): print("true") else: print("false")

C#

// C# program to search an element in row-wise // and column-wise sorted matrix

using System;

class GfG { static bool matSearch(int[][] mat, int x) { int n = mat.Length, m = mat[0].Length;

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (mat[i][j] == x)
                return true;
        }
    }

    // If x was not found, return false
    return false;
}

static void Main() {
    int[][] mat = new int[][] {
        new int[] {3, 30, 38},
        new int[] {20, 52, 54},
        new int[] {35, 60, 69}
    };
    int x = 35;

    if (matSearch(mat, x))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

// Java Script program to search an element in row-wise // and column-wise sorted matrix

function matSearch(mat, x) { const n = mat.length, m = mat[0].length;

for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
        if (mat[i][j] === x)
            return true;
    }
}

// If x was not found, return false
return false;

}

// Driver Code const mat = [ [ 3, 30, 38 ], [ 20, 52, 54 ], [ 35, 60, 69 ] ]; const x = 35;

if (matSearch(mat, x)) console.log("true"); else console.log("false");

`

[Better Approach] Binary Search - O(n*logm) Time and O(1) Space:

To optimize the above approach we are going to use the **Binary Search algorithm.
The problem specifies that each **row in the given matrix is **sorted in ascending order. Instead of searching each **column sequentially, we can efficiently apply **Binary Search on each **row to determine if the **target is present.

C++ `

// C++ program to search an element in row-wise // and column-wise sorted matrix

#include #include

using namespace std;

bool binarySearch(vector &mat, int target) { int n = mat.size(); int low = 0, high = n - 1;

// Standard binary search algorithm
while (low <= high) {
    int mid = (low + high) / 2;
    
    if (mat[mid] == target) 
        return true;  // Element found
    else if (target > mat[mid]) 
        low = mid + 1; // Search in the right half
    else 
        high = mid - 1; // Search in the left half
}
return false;  // Element not found

}

bool matSearch(vector<vector> &mat, int x) { int n = mat.size();

// Iterate over each row and perform binary search
for (int i = 0; i < n; i++) {
    if (binarySearch(mat[i], x)) 
        return true;  // Element found in one of the rows
}

return false;  // Element not found in any row

}

int main() { vector<vector> mat = {{3, 30, 38}, {20, 52, 54}, {35, 60, 69}}; int x = 35; if(matSearch(mat, x)) cout << "true"; else cout << "false"; return 0; }

Java

// Java program to search an element in row-wise // and column-wise sorted matrix

class GfG { public static boolean binarySearch(int[] mat, int target) { int low = 0, high = mat.length - 1;

    // Standard binary search algorithm
    while (low <= high) {
        int mid = (low + high) / 2;

        if (mat[mid] == target) 
            return true;  // Element found
        else if (target > mat[mid]) 
            low = mid + 1; // Search in the right half
        else 
            high = mid - 1; // Search in the left half
    }
    return false;  // Element not found
}

static boolean matSearch(int[][] mat, int x) {
    int n = mat.length; // Number of rows

    // Iterate over each row and perform binary search
    for (int i = 0; i < n; i++) {
        if (binarySearch(mat[i], x)) 
            return true;  // Element found in one of the rows
    }
    return false;  // Element not found in any row
}

public static void main(String[] args) {
    int[][] mat = {{3, 30, 38},
                   {20, 52, 54},
                   {35, 60, 69}};
    int x = 35;
    
    if(matSearch(mat, x)) 
        System.out.println("true");
    else 
        System.out.println("false");
}

}

Python

Python program to search an element in row-wise

and column-wise sorted matrix

def binarySearch(mat, target):

n = len(mat)
low, high = 0, n - 1

# Standard binary search algorithm
while low <= high:
    mid = (low + high) // 2  # Midpoint index

    if mat[mid] == target:
        return True  # Element found
    elif target > mat[mid]:
        low = mid + 1  # Search in the right half
    else:
        high = mid - 1  # Search in the left half

return False  # Element not found

def matSearch(mat, x):

n = len(mat)
m = len(mat[0])

# Iterate over each row and perform binary search
for i in range(n):
    if binarySearch(mat[i], x):
        return True  # Element found in one of the rows

return False  # Element not found in any row

if name == "main": mat = [ [3, 30, 38], [20, 52, 54], [35, 60, 69] ] x = 35 if matSearch(mat, x): print("true") else: print("false")

C#

// C# program to search an element in row-wise // and column-wise sorted matrix

using System;

class GfG {

// Function to perform binary search on a sorted row (1D array)
static bool BinarySearch(int[] mat, int target)
{
    int low = 0, high = mat.Length - 1;

    // Standard binary search algorithm
    while (low <= high)
    {
        int mid = (low + high) / 2;

        if (mat[mid] == target)
            return true; // Element found
        else if (target > mat[mid])
            low = mid + 1; // Search in the right half
        else
            high = mid - 1; // Search in the left half
    }
    return false; // Element not found
}

// Function to search an element in a row-wise sorted matrix
static bool matSearch(int[][] mat, int x) {
    int n = mat.Length;

    // Iterate over each row and perform binary search
    for (int i = 0; i < n; i++)
    {
        if (BinarySearch(mat[i], x))
            return true; // Element found in one of the rows
    }

    return false; // Element not found in any row
}

static void Main() {
    int[][] mat = new int[][] {
        new int[] {3, 30, 38},
        new int[] {20, 52, 54},
        new int[] {35, 60, 69}
    };
    int x = 35;

    if (matSearch(mat, x))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

// JavaScript program to search an element in row-wise // and column-wise sorted matrix

function binarySearch(mat, target) { let low = 0, high = mat.length - 1;

// Standard binary search algorithm
while (low <= high) {
    let mid = Math.floor((low + high) / 2); // Use Math.floor() to get an integer index

    if (mat[mid] === target) 
        return true;  // Element found
    else if (target > mat[mid]) 
        low = mid + 1; // Search in the right half
    else 
        high = mid - 1; // Search in the left half
}
return false;  // Element not found

}

function matSearch(mat, x) { let n = mat.length;

// Iterate over each row and perform binary search
for (let i = 0; i < n; i++) {
    if (binarySearch(mat[i], x)) 
        return true;  // Element found in one of the rows
}

return false;  // Element not found in any row

}

// Driver Code let mat = [ [3, 30, 38], [20, 52, 54], [35, 60, 69] ]; let x = 35;

if (matSearch(mat, x)) console.log("true"); else console.log("false");

`

[Expected Approach] Eliminating rows or columns - O(n + m) Time and O(1) Space:

The idea is to remove a row or column in each comparison until an element is found. Start searching from the top-right corner of the matrix. There are 3 possible cases:

  1. **x is greater than the current element: This ensures that all the elements in the current row are smaller than the given number as the pointer is already at the right-most element and the row is sorted. Thus, the entire row gets eliminated and continues the search from the next row.
  2. **x is smaller than the current element: This ensures that all the elements in the current column are greater than the given number. Thus, the entire column gets eliminated and continues the search from the previous column, i.e. the column on the immediate left.
  3. **The given number is equal to the current number: This will end the search.

**Illustration:

C++ `

// C++ program to search an element in row-wise // and column-wise sorted matrix

#include #include

using namespace std;

bool matSearch(vector<vector> &mat, int x) { int n = mat.size(), m = mat[0].size(); int i = 0, j = m - 1;

while(i < n && j >= 0) {
  
    // If x > mat[i][j], then x will be greater
    // than all elements to the left of 
    // mat[i][j] in row i, so increment i
    if(x > mat[i][j]) {
        i++;
    }
  
    // If x < mat[i][j], then x will be smaller
    // than all elements to the bottom of
    // mat[i][j] in column j, so decrement j
    else if(x < mat[i][j]) {
        j--;
    }
  
    // If x = mat[i][j], return true
    else {
        return true;
    }
}

// If x was not found, return false
return false;

}

int main() { vector<vector> mat = {{3, 30, 38}, {20, 52, 54}, {35, 60, 69}}; int x = 35; if(matSearch(mat, x)) cout << "true"; else cout << "false"; return 0; }

Java

// Java program to search an element in row-wise // and column-wise sorted matrix

import java.util.*;

class GfG { static boolean matSearch(int[][] mat, int x) { int n = mat.length, m = mat[0].length; int i = 0, j = m - 1;

    while (i < n && j >= 0) {

        // If x > mat[i][j], then x will be greater
        // than all elements to the left of 
        // mat[i][j] in row i, so increment i
        if (x > mat[i][j]) {
            i++;
        }

        // If x < mat[i][j], then x will be smaller
        // than all elements to the bottom of
        // mat[i][j] in column j, so decrement j
        else if (x < mat[i][j]) {
            j--;
        }

        // If x = mat[i][j], return true
        else {
            return true;
        }
    }

    // If x was not found, return false
    return false;
}

public static void main(String[] args) {
    int[][] mat = {
        {3, 30, 38},
        {20, 52, 54},
        {35, 60, 69}
    };
    int x = 35;

    if (matSearch(mat, x))
        System.out.println("true");
    else
        System.out.println("false");
}

}

Python

Python program to search an element in row-wise

and column-wise sorted matrix

def matSearch(mat, x): n = len(mat) m = len(mat[0]) i = 0 j = m - 1

while i < n and j >= 0:
  
    # If x > mat[i][j], then x will be greater
    # than all elements to the left of 
    # mat[i][j] in row i, so increment i
    if x > mat[i][j]:
        i += 1
  
    # If x < mat[i][j], then x will be smaller
    # than all elements to the bottom of
    # mat[i][j] in column j, so decrement j
    elif x < mat[i][j]:
        j -= 1
  
    # If x = mat[i][j], return true
    else:
        return True

# If x was not found, return false
return False

if name == "main": mat = [ [3, 30, 38], [20, 52, 54], [35, 60, 69] ] x = 35 if matSearch(mat, x): print("true") else: print("false")

C#

// C# program to search an element in row-wise // and column-wise sorted matrix

using System;

class GfG { static bool matSearch(int[][] mat, int x) { int n = mat.Length, m = mat[0].Length; int i = 0, j = m - 1;

    while (i < n && j >= 0) {

        // If x > mat[i][j], then x will be greater
        // than all elements to the left of 
        // mat[i][j] in row i, so increment i
        if (x > mat[i][j]) {
            i++;
        }

        // If x < mat[i][j], then x will be smaller
        // than all elements to the bottom of
        // mat[i][j] in column j, so decrement j
        else if (x < mat[i][j]) {
            j--;
        }

        // If x = mat[i][j], return true
        else {
            return true;
        }
    }

    // If x was not found, return false
    return false;
}

static void Main() {
    int[][] mat = new int[][] {
        new int[] {3, 30, 38},
        new int[] {20, 52, 54},
        new int[] {35, 60, 69}
    };
    int x = 35;

    if (matSearch(mat, x))
        Console.WriteLine("true");
    else
        Console.WriteLine("false");
}

}

JavaScript

// JavaScript program to search an element in row-wise // and column-wise sorted matrix

function matSearch(mat, x) { let n = mat.length, m = mat[0].length; let i = 0, j = m - 1;

while (i < n && j >= 0) {

    // If x > mat[i][j], then x will be greater
    // than all elements to the left of 
    // mat[i][j] in row i, so increment i
    if (x > mat[i][j]) {
        i++;
    }

    // If x < mat[i][j], then x will be smaller
    // than all elements to the bottom of
    // mat[i][j] in column j, so decrement j
    else if (x < mat[i][j]) {
        j--;
    }

    // If x = mat[i][j], return true
    else {
        return true;
    }
}

// If x was not found, return false
return false;

}

// Driver Code let mat = [ [3, 30, 38], [20, 52, 54], [35, 60, 69] ]; let x = 35;

if (matSearch(mat, x)) console.log("true"); else console.log("false");

`

**Related Article: Search element in a sorted matrix