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.

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 implementation of above algorithm:

C++

 // Finding peak element in a 2D Array. #include 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 - ceil((double)mid / 2));        // If max is less than its left     // if (max < arr[max_index][mid+1])     return findPeakRec(arr, rows, columns, mid + ceil((double)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; }

Java

 // Finding peak element in a 2D Array. class GFG  {     static int MAX = 100;        // Function to find the maximum in column      // 'mid', 'rows' is number of rows.     static int findMax(int[][] arr, 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 change the value of [max]     static int Max(int[][] arr, int rows,                    int mid, int max)     {         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];             }         }         return max;     }        // Function to find a peak element     static int findPeakRec(int[][] arr, 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);         max = Max(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,                           (int)(mid - Math.ceil((double) mid / 2)));            // If max is less than its left         // if (max < arr[max_index][mid+1])         return findPeakRec(arr, rows, columns,                       (int)(mid + Math.ceil((double) mid / 2)));     }        // A wrapper over findPeakRec()     static int findPeak(int[][] arr, int rows, int columns)      {         return findPeakRec(arr, rows, columns, columns / 2);     }        // Driver Code     public static void main(String[] args)     {         int[][] arr = {{ 10, 8, 10, 10 },                        { 14, 13, 12, 11 },                         { 15, 9, 11, 21 },                         { 16, 17, 19, 20 }};                    // Number of Columns         int rows = 4, columns = 4;         System.out.println(findPeak(arr, rows, columns));     } }    // This code is contributed by // sanjeev2552

Python3

 # Finding peak element in a 2D Array. MAX = 100 from math import ceil    # Function to find the maximum in column 'mid' # 'rows' is number of rows. def findMax(arr, rows, mid,max):        max_index = 0     for i in range(rows):         if (max < arr[i][mid]):                            # Saving global maximum and its index             # to check its neighbours             max = arr[i][mid]             max_index = i     #print(max_index)        return max,max_index    # Function to find a peak element def findPeakRec(arr, rows, columns,mid):        # Evaluating maximum of mid column.      # Note max is passed by reference.     max = 0     max, max_index = findMax(arr, rows, mid, max)        # If we are on the first or last column,     # max is a peak     if (mid == 0 or mid == columns - 1):         return max        # If mid column maximum is also peak     if (max >= arr[max_index][mid - 1] and          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 - ceil(mid / 2.0))        # If max is less than its left     # if (max < arr[max_index][mid+1])     return findPeakRec(arr, rows, columns,                         mid + ceil(mid / 2.0))    # A wrapper over findPeakRec() def findPeak(arr, rows, columns):     return findPeakRec(arr, rows,                         columns, columns // 2)    # Driver Code arr = [ [ 10, 8, 10, 10 ],         [ 14, 13, 12, 11 ],         [ 15, 9, 11, 21 ],         [ 16, 17, 19, 20 ] ]    # Number of Columns rows = 4 columns = 4 print(findPeak(arr, rows, columns))    # This code is contributed by Mohit Kumar

C#

 // Finding peak element in a 2D Array. using System;    class GFG {        // Function to find the maximum in column      // 'mid', 'rows' is number of rows.     static int findMax(int[,] arr, 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 change the value of [max]     static int Max(int[,] arr, int rows,                    int mid, int max)     {         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];             }         }         return max;     }        // Function to find a peak element     static int findPeakRec(int[,] arr, 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);         max = Max(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,                     (int)(mid - Math.Ceiling((double) mid / 2)));            // If max is less than its left         // if (max < arr[max_index][mid+1])         return findPeakRec(arr, rows, columns,                 (int)(mid + Math.Ceiling((double) mid / 2)));     }        // A wrapper over findPeakRec()     static int findPeak(int[,] arr,                          int rows, int columns)      {         return findPeakRec(arr, rows, columns,                                        columns / 2);     }        // Driver Code     static public void Main ()     {         int[,] arr = {{ 10, 8, 10, 10 },                       { 14, 13, 12, 11 },                        { 15, 9, 11, 21 },                        { 16, 17, 19, 20 }};                // Number of Columns         int rows = 4, columns = 4;         Console.Write(findPeak(arr, rows, columns));     } }    // This code is contributed by ajit.

Output:

21

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

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

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