Find a peak element in a 2D array

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.


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.

  1. Consider mid column and find maximum element in it.
  2. 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]’.
  3. If max >= A[index][mid-1] & max >= A[index][pick+1], max is a peak, return max.
  4. If max < mat[max_index][mid-1], recur for left half of matrix.
  5. 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.
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.

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