Given a sorted matrix mat[n][m] and an element ‘x’. Find position of x in the matrix if it is present, else print -1. 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, first element of row 'i' is greater than or equal to the last element of row 'i-1'. The approach should have O(log n + log m) time complexity.

Examples:

Input : mat[][] = { {1, 5, 9}, {14, 20, 21}, {30, 34, 43} } x = 14 Output : Found at (1, 0) Input : mat[][] = { {1, 5, 9, 11}, {14, 20, 21, 26}, {30, 34, 43, 50} } x = 42 Output : -1

Please note that this problem is different from Search in a row wise and column wise sorted matrix. Here matrix is more strictly sorted as first element of a row is greater than last element of previous row.

A **Simple Solution **is to one by one compare x with every element of matrix. If matches, then return position. If we reach end, return -1. Time complexity of this solution is O(n x m).

An **efficient solution** is to typecast given 2D array to 1D array, then apply binary search on the typecasted array.

**Another efficient approach** that doesn’t require typecasting is explained below.

1) Perform binary search on the middle column till only two elements are left or till the middle element of some row in the search is the required element 'x'. This search is done to skip the rows that are not required 2) The two left elements must be adjacent. Consider the rows of two elements and do following a) check whether the element 'x' equals to the middle element of any one of the 2 rows b) otherwise according to the value of the element 'x' check whether it is present in the 1st half of 1st row, 2nd half of 1st row, 1st half of 2nd row or 2nd half of 2nd row. Note: This approach works for the matrix n x m where 2 <= n. The algorithm can be modified for matrix 1 x m, we just need to check whether 2nd row exists or not

__Example:__

Consider: | 123 4| x = 3, mat = | 567 8| Middle column: | 91011 12| = {2, 6, 10, 14} |131415 16| perform binary search on them since, x < 6, discard the last 2 rows as 'a' will not lie in them(sorted matrix) Now, only two rows are left | 123 4| x = 3, mat = | 567 8| Check whether element is present on the middle elements of these rows = {2, 6} x != 2 or 6 If not, consider the four sub-parts1st half of 1st row= {1},2nd half of 1st row= {3, 4}1st half of 2nd row= {5},2nd half of 2nd row= {7, 8} According the value of'x'it will be searched in the 2nd half of 1st row ={3, 4}and found at(i, j):(0, 2)

## C++

// C++ implementation to search an element in a // sorted matrix #include <bits/stdc++.h> using namespace std; const int MAX = 100; // This function does Binary search for x in i-th // row. It does the search from mat[i][j_low] to // mat[i][j_high] void binarySearch(int mat[][MAX], int i, int j_low, int j_high, int x) { while (j_low <= j_high) { int j_mid = (j_low + j_high) / 2; // Element found if (mat[i][j_mid] == x) { cout << "Found at (" << i << ", " << j_mid << ")"; return; } else if (mat[i][j_mid] > x) j_high = j_mid - 1; else j_low = j_mid + 1; } // element not found cout << "Element no found"; } // Function to perform binary search on the mid // values of row to get the desired pair of rows // where the element can be found void sortedMatrixSearch(int mat[][MAX], int n, int m, int x) { // Single row matrix if (n == 1) { binarySearch(mat, 0, 0, m-1, x); return; } // Do binary search in middle column. // Condition to terminate the loop when the // 2 desired rows are found int i_low = 0; int i_high = n-1; int j_mid = m/2; while ((i_low+1) < i_high) { int i_mid = (i_low + i_high) / 2; // element found if (mat[i_mid][j_mid] == x) { cout << "Found at (" << i_mid << ", " << j_mid << ")"; return; } else if (mat[i_mid][j_mid] > x) i_high = i_mid; else i_low = i_mid; } // If element is present on the mid of the // two rows if (mat[i_low][j_mid] == x) cout << "Found at (" << i_low << "," << j_mid << ")"; else if (mat[i_low+1][j_mid] == x) cout << "Found at (" << (i_low+1) << ", " << j_mid << ")"; // Ssearch element on 1st half of 1st row else if (x <= mat[i_low][j_mid-1]) binarySearch(mat, i_low, 0, j_mid-1, x); // Search element on 2nd half of 1st row else if (x >= mat[i_low][j_mid+1] && x <= mat[i_low][m-1]) binarySearch(mat, i_low, j_mid+1, m-1, x); // Search element on 1st half of 2nd row else if (x <= mat[i_low+1][j_mid-1]) binarySearch(mat, i_low+1, 0, j_mid-1, x); // search element on 2nd half of 2nd row else binarySearch(mat, i_low+1, j_mid+1, m-1, x); } // Driver program to test above int main() { int n = 4, m = 5, x = 8; int mat[][MAX] = {{0, 6, 8, 9, 11}, {20, 22, 28, 29, 31}, {36, 38, 50, 61, 63}, {64, 66, 100, 122, 128}}; sortedMatrixSearch(mat, n, m, x); return 0; }

## Java

// java implementation to search // an element in a sorted matrix import java.io.*; class GFG { static int MAX = 100; // This function does Binary search for x in i-th // row. It does the search from mat[i][j_low] to // mat[i][j_high] static void binarySearch(int mat[][], int i, int j_low, int j_high, int x) { while (j_low <= j_high) { int j_mid = (j_low + j_high) / 2; // Element found if (mat[i][j_mid] == x) { System.out.println ( "Found at (" + i + ", " + j_mid +")"); return; } else if (mat[i][j_mid] > x) j_high = j_mid - 1; else j_low = j_mid + 1; } // element not found System.out.println ( "Element no found"); } // Function to perform binary search on the mid // values of row to get the desired pair of rows // where the element can be found static void sortedMatrixSearch(int mat[][], int n, int m, int x) { // Single row matrix if (n == 1) { binarySearch(mat, 0, 0, m - 1, x); return; } // Do binary search in middle column. // Condition to terminate the loop when the // 2 desired rows are found int i_low = 0; int i_high = n - 1; int j_mid = m / 2; while ((i_low + 1) < i_high) { int i_mid = (i_low + i_high) / 2; // element found if (mat[i_mid][j_mid] == x) { System.out.println ( "Found at (" + i_mid +", " + j_mid +")"); return; } else if (mat[i_mid][j_mid] > x) i_high = i_mid; else i_low = i_mid; } // If element is present on // the mid of the two rows if (mat[i_low][j_mid] == x) System.out.println ( "Found at (" + i_low + "," + j_mid +")"); else if (mat[i_low + 1][j_mid] == x) System.out.println ( "Found at (" + (i_low + 1) + ", " + j_mid +")"); // Ssearch element on 1st half of 1st row else if (x <= mat[i_low][j_mid - 1]) binarySearch(mat, i_low, 0, j_mid - 1, x); // Search element on 2nd half of 1st row else if (x >= mat[i_low][j_mid + 1] && x <= mat[i_low][m - 1]) binarySearch(mat, i_low, j_mid + 1, m - 1, x); // Search element on 1st half of 2nd row else if (x <= mat[i_low + 1][j_mid - 1]) binarySearch(mat, i_low + 1, 0, j_mid - 1, x); // search element on 2nd half of 2nd row else binarySearch(mat, i_low + 1, j_mid + 1, m - 1, x); } // Driver program public static void main (String[] args) { int n = 4, m = 5, x = 8; int mat[][] = {{0, 6, 8, 9, 11}, {20, 22, 28, 29, 31}, {36, 38, 50, 61, 63}, {64, 66, 100, 122, 128}}; sortedMatrixSearch(mat, n, m, x); } } // This code is contributed by vt_m

Output:

Found at (0,2)

Time complexity: O(log n + log m). O(Log n) time is required to find the two desired rows. Then O(Log m) time is required for binary search in one of the four parts with size equal to m/2.

This article is contributed by **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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