An element is a peak element if it is greater than or equal to its four neighbors, left, right, top and bottom. For example neighbors for A[i][j] are A[i-1][j], A[i+1][j], A[i][j-1] and A[i][j+1]. For corner elements, missing neighbors are considered of negative infinite value.

Examples:

Input : 10 20 15 21 30 14 7 16 32 Output : 30 30 is a peak element because all its neighbors are smaller or equal to it. 32 can also be picked as a peak. Input : 10 7 11 17 Output : 17

Below are some facts about this problem:

1: A Diagonal adjacent is not considered as neighbor.

2: A peak element is not necessarily the maximal element.

3: More than one such elements can exist.

4: There is always a peak element. We can see this property by creating some matrices using pen and paper.

**Method 1: (Brute Force)**

Iterate through all the elements of Matrix and check if it is greater/equal to all its neighbors. If yes, return the element.

Time Complexity: O(rows * columns)

Auxiliary Space: O(1)

**Method 2 : (Efficient)**

This problem is mainly an extension of Find a peak element in 1D array. We apply similar Binary Search based solution here.

- Consider mid column and find maximum element in it.
- Let index of mid column be ‘mid’, value of maximum element in mid column be ‘max’ and maximum element be at ‘mat[max_index][mid]’.
- If max >= A[index][mid-1] & max >= A[index][pick+1], max is a peak, return max.
- If max < mat[max_index][mid-1], recur for left half of matrix.
- If max < mat[max_index][mid+1], recur for right half of matrix.

Below is the C++ implementation of above algorithm:

// Finding peak element in a 2D Array. #include<bits/stdc++.h> using namespace std; const int MAX = 100; // Function to find the maximum in column 'mid' // 'rows' is number of rows. int findMax(int arr[][MAX], int rows, int mid, int &max) { int max_index = 0; for (int i = 0; i < rows; i++) { if (max < arr[i][mid]) { // Saving global maximum and its index // to check its neighbours max = arr[i][mid]; max_index = i; } } return max_index; } // Function to find a peak element int findPeakRec(int arr[][MAX], int rows, int columns, int mid) { // Evaluating maximum of mid column. Note max is // passed by reference. int max = 0; int max_index = findMax(arr, rows, mid, max); // If we are on the first or last column, // max is a peak if (mid == 0 || mid == columns-1) return max; // If mid column maximum is also peak if (max >= arr[max_index][mid-1] && max >= arr[max_index][mid+1]) return max; // If max is less than its left if (max < arr[max_index][mid-1]) return findPeakRec(arr, rows, columns, mid - mid/2); // If max is less than its left // if (max < arr[max_index][mid+1]) return findPeakRec(arr, rows, columns, mid+mid/2); } // A wrapper over findPeakRec() int findPeak(int arr[][MAX], int rows, int columns) { return findPeakRec(arr, rows, columns, columns/2); } // Driver Code int main() { int arr[][MAX] = {{ 10, 8, 10, 10 }, { 14, 13, 12, 11 }, { 15, 9, 11, 21 }, { 16, 17, 19, 20 } }; // Number of Columns int rows = 4, columns = 4; cout << findPeak(arr, rows, columns); return 0; }

Output: 21

Time Complexity : O(rows * log(columns)). We recur for half number of columns. In every recursive call we linearly search for maximum in current mid column.

Auxiliary Space : O(columns/2) for Recursion Call Stack

This article is contributed by **Rohit Thapliyal**. 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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