# Search element in a sorted matrix

Given a sorted matrix mat[n][m] and an element ‘x’. Find the 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, the 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 the first element of a row is greater than the last element of the previous row.
A Simple Solution is to one by one compare x with every element of the matrix. If matches, then return position. If we reach the end, return -1. The time complexity of this solution is O(n x m).
An efficient solution is to typecast a given 2D array to a 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:    | 1  2  3  4|
x = 3, mat = | 5  6  7  8|   Middle column:
| 9 10 11 12|    = {2, 6, 10, 14}
|13 14 15 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
| 1  2  3  4|
x = 3, mat = | 5  6  7  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-parts
1st 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++ implementation to search an element in a // sorted matrix #include 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 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

 # Python3 implementation # to search an element in a # sorted matrix 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] def binarySearch(mat, i, j_low,                  j_high, x):       while (j_low <= j_high):               j_mid = (j_low + j_high) // 2           # Element found         if (mat[i][j_mid] == x):                       print("Found at (", i, ", ", j_mid, ")")             return           elif (mat[i][j_mid] > x):             j_high = j_mid - 1           else:             j_low = j_mid + 1          # Element not found     print ("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 def sortedMatrixSearch(mat, n, m, 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     i_low = 0     i_high = n - 1     j_mid = m // 2     while ((i_low + 1) < i_high):               i_mid = (i_low + i_high) // 2           # element found         if (mat[i_mid][j_mid] == x):                       print ("Found at (", i_mid, ", ", j_mid, ")")             return           elif (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):         print ("Found at (" , i_low, ",", j_mid , ")")     elif (mat[i_low + 1][j_mid] == x):         print ("Found at (", (i_low + 1), ", ", j_mid, ")")       # Ssearch element on 1st half of 1st row     elif (x <= mat[i_low][j_mid - 1]):         binarySearch(mat, i_low, 0, j_mid - 1, x)       # Search element on 2nd half of 1st row     elif (x >= mat[i_low][j_mid + 1] and           x <= mat[i_low][m - 1]):        binarySearch(mat, i_low, j_mid + 1, m - 1, x)       # Search element on 1st half of 2nd row     elif (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 if __name__ == "__main__":       n = 4     m = 5     x = 8     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 Chitranayal

 // C# implementation to search // an element in a sorted matrix using System;   class GFG {     // 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)             {                 Console.Write ( "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         Console.Write ( "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)             {                                   Console.Write ( "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)         Console.Write ( "Found at (" + i_low +                            "," + j_mid +")");         else if (mat[i_low + 1,j_mid] == x)         Console.Write ( "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 parashar...

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.
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