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. The task is to find the index of the peak element.

Examples:

Input: 10 20 15
          21 30 14
         7  16 32 
Output: 1, 1
Explanation: The value at index {1, 1} is 30, which is a peak element because all its neighbors are smaller or equal to it. Similarly, {2, 2} can also be picked as a peak.

Input: 10 7
          11 17
Output : 1, 1

Below are some facts about this problem: 

1.  A Diagonal adjacent is not considered a neighbour. 
2. A peak element is not necessarily the maximal element. 
3. More than one such element 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 neighbours. If yes, return the element.

C++

// Finding peak element in a 2D Array.
#include <bits/stdc++.h>
using namespace std;

vector<int> findPeakGrid(vector<vector<int>> arr){
    vector<int> result;
    int row = arr.size();
    int column = arr[0].size();

    for(int i = 0; i<row; i++){
        for(int j = 0; j<column; j++){
            // checking with top element
            if(i > 0)
                if(arr[i][j] < arr[i-1][j]) continue;
            // checking with right element
            if(j < column-1)
                if(arr[i][j] < arr[i][j+1]) continue;
            // checking with bottom element
            if(i < row-1)
                if(arr[i][j] < arr[i+1][j]) continue;
            // checking with left element
            if(j > 0)
                if(arr[i][j] < arr[i][j-1])  continue;

            result.push_back(i);
            result.push_back(j);
            break;
        }
    }
    return result;
}

// Driver Code
int main(){
    vector<vector<int>> arr = {{9,8}, {2,6}};
    vector<int> result = findPeakGrid(arr);
    cout<<"Peak element found at index: "<<result[0]<<", "<<result[1]<<endl;
    return 0;
}

// This code is contributed by Yash Agarwal(yashagarwal2852002)

Java

import java.util.*;

public class Main {
    public static List<Integer> findPeakGrid(int[][] arr)
    {
        List<Integer> result = new ArrayList<>();
        int row = arr.length;
        int column = arr[0].length;

        for (int i = 0; i < row; i++) {
            for (int j = 0; j < column; j++) {
                // checking with top element
                if (i > 0)
                    if (arr[i][j] < arr[i - 1][j])
                        continue;
                // checking with right element
                if (j < column - 1)
                    if (arr[i][j] < arr[i][j + 1])
                        continue;
                // checking with bottom element
                if (i < row - 1)
                    if (arr[i][j] < arr[i + 1][j])
                        continue;
                // checking with left element
                if (j > 0)
                    if (arr[i][j] < arr[i][j - 1])
                        continue;

                result.add(i);
                result.add(j);
                break;
            }
        }
        return result;
    }

    public static void main(String[] args)
    {
        int[][] arr = { { 9, 8 }, { 2, 6 } };
        List<Integer> result = findPeakGrid(arr);
        System.out.println("Peak element found at index: "
                           + result.get(0) + ", "
                           + result.get(1));
    }
}

Python3

# Finding a peak element in 2D array
def findPeakGrid(arr):
    result = []
    row = len(arr)
    column = len(arr[0])

    for i in range(row):
        for j in range(column):

            # checking with top element
            if i > 0:
                if arr[i][j] < arr[i-1][j]:
                    continue
            # checking with right element
            if j < column-1:
                if arr[i][j] < arr[i][j+1]:
                    continue
            # checking with bottom element
            if i < row-1:
                if arr[i][j] < arr[i+1][j]:
                    continue
            # checking with left element
            if j > 0:
                if arr[i][j] < arr[i][j-1]:
                    continue

            result.append(i)
            result.append(j)
            break

    return result

# driver code
arr = [[9,8], [2,6]]
result = findPeakGrid(arr)
print("Peak element found at index:", result)

# This code is constributed by phasing17

C#

// C# code to find peak element in a 2D array
using System;
using System.Collections.Generic;
class GFG {
  static int[] findPeakGrid(int[][] arr){
    int[] result = new int[2];
    int row = arr.Length;
    int column = arr[0].Length;
    for(int i = 0; i<row; i++){
      for(int j = 0; j<column; j++)
      {

        // checking with top element
        if(i > 0)
          if(arr[i][j] < arr[i-1][j]) continue;

        // checking with right element
        if(j < column-1)
          if(arr[i][j] < arr[i][j+1]) continue;

        // checking with bottom element
        if(i < row-1)
          if(arr[i][j] < arr[i+1][j]) continue;

        // checking with left element
        if(j > 0)
          if(arr[i][j] < arr[i][j-1])  continue;
        result[0] = i;
        result[1] = j;
        break;
      }
    }
    return result;
  }

  // driver code to test above function
  public static void Main(){
    int[][] arr = { new[] { 9, 8 }, new[] { 2, 6 } };
    int[] result = findPeakGrid(arr);
    Console.WriteLine("Peak element found at index: "
                      + result[0] + "," + result[1]);
  }
}

// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGAWRAL2852002)

Javascript

// Finding a peak element in 2D array
function findPeakGrid(arr){
    let result = [];
    let row = arr.length;
    let column = arr[0].length;

    for(let i = 0; i<row; i++){
        for(let j = 0; j<column; j++){
            // checking with top element
            if(i > 0)
                if(arr[i][j] < arr[i-1][j]) continue;
            // checking with right element
            if(j < column-1)
                if(arr[i][j] < arr[i][j+1]) continue;
            // checking with bottom element
            if(i < row-1)
                if(arr[i][j] < arr[i+1][j]) continue;
            // checking with left element
            if(j > 0)
                if(arr[i][j] < arr[i][j-1])  continue;

            result.push(i);
            result.push(j);
            break;
        }
    }
    return result;
}

// driver code
let arr = [[9,8], [2,6]];
let result = findPeakGrid(arr);
console.log("Peak element found at index: " + result[0] + ", " + result[1]);
// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL(KIRTIAGARWAL23121999)

Output

Peak element found at index: 0, 0

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][mid+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 the above algorithm: 

C++

// Finding peak element in a 2D Array.
#include <bits/stdc++.h>
using namespace std;

// Finding peak element in a 2D Array.
#include <bits/stdc++.h>
using namespace std;

vector<int> findPeakGrid(vector<vector<int> >& mat)
{

    int stcol = 0,
        endcol
        = mat[0].size()
          - 1; // Starting point & end point of Search Space

    while (stcol <= endcol) { // Bin Search Condition

        int midcol = stcol + (endcol - stcol) / 2,
            ansrow = 0;
        // "ansrow" To keep the row number of global Peak
        // element of a column

        // Finding the row number of Global Peak element in
        // Mid Column.
        for (int r = 0; r < mat.size(); r++) {
            ansrow = mat[r][midcol] >= mat[ansrow][midcol]
                         ? r
                         : ansrow;
        }

        // Finding next Search space will be left or right
        bool valid_left = midcol - 1 >= stcol
                          && mat[ansrow][midcol - 1]
                                 > mat[ansrow][midcol];
        bool valid_right = midcol + 1 <= endcol
                           && mat[ansrow][midcol + 1]
                                  > mat[ansrow][midcol];

        // if we're at Peak Element
        if (!valid_left && !valid_right) {
            return { ansrow, midcol };
        }

        else if (valid_right)
            stcol = midcol
                    + 1; // move the search space in right
        else
            endcol = midcol
                     - 1; // move the search space in left
    }

    return { -1, -1 };
}


// Driver Code
int main()
{
    vector<vector<int> > arr = {{9, 8}, {2 ,6}};

    vector<int> result = findPeakGrid(arr);

    cout << "Peak element found at index: " << result[0] << "," << result[1] << endl;
    return 0;
}

Java

// Finding peak element in a 2D Array.
import java.util.*;

public class GFG {

  static int[] findPeakGrid(int[][] mat)
  {

    // Starting point & end point of Search Space
    int stcol = 0, endcol = mat[0].length - 1;

    // Bin Search Condition
    while (stcol <= endcol) {

      int midcol = stcol + (endcol - stcol) / 2,
      ansrow = 0;
      // "ansrow" To keep the row number of global
      // Peak element of a column

      // Finding the row number of Global Peak element
      // in Mid Column.
      for (int r = 0; r < mat.length; r++) {
        ansrow
          = mat[r][midcol] >= mat[ansrow][midcol]
          ? r
          : ansrow;
      }

      // Finding next Search space will be left or
      // right
      boolean valid_left
        = midcol - 1 >= stcol
        && mat[ansrow][midcol - 1]
        > mat[ansrow][midcol];
      boolean valid_right
        = midcol + 1 <= endcol
        && mat[ansrow][midcol + 1]
        > mat[ansrow][midcol];

      // if we're at Peak Element
      if (!valid_left && !valid_right) {
        return new int[] { ansrow, midcol };
      }

      else if (valid_right)
        stcol
        = midcol
        + 1; // move the search space in right
      else
        endcol
        = midcol
        - 1; // move the search space in left
    }

    return new int[] { -1, -1 };
  }

  // Driver Code
  public static void main(String[] args)
  {
    int[][] arr = { { 9, 8 }, { 2, 6 } };

    int[] result = findPeakGrid(arr);

    System.out.println("Peak element found at index: "
                       + result[0] + "," + result[1]);
  }
}

// This code is contributed by Karandeep1234

Python3

# Finding peak element in a 2D Array.
def findPeakGrid(mat):
    stcol = 0
    endcol = len(mat[0]) - 1; # Starting po  end po of Search Space

    while (stcol <= endcol):  # Bin Search Condition

        midcol = stcol + int((endcol - stcol) / 2)
        ansrow = 0;
        # "ansrow" To keep the row number of global Peak
        # element of a column

        # Finding the row number of Global Peak element in
        # Mid Column.
        for r in range(len(mat)):
            ansrow = r if mat[r][midcol] >= mat[ansrow][midcol] else ansrow;


        # Finding next Search space will be left or right
        valid_left =  midcol - 1 >= stcol and mat[ansrow][midcol - 1] > mat[ansrow][midcol];
        valid_right = midcol + 1 <= endcol and mat[ansrow][midcol + 1] > mat[ansrow][midcol];

        # if we're at Peak Element
        if (not valid_left and not valid_right) :
            return [ ansrow, midcol ];


        elif (valid_right):
            stcol = midcol  + 1; # move the search space in right
        else:
            endcol = midcol  - 1; # move the search space in left

    return [ -1, -1 ];

# Driver Code
arr = [[9, 8], [2 ,6]];
result = findPeakGrid(arr);
print("Peak element found at index:", result)

# This code is contributed by phasing17.

C#

// Finding peak element in a 2D Array.
using System;
using System.Collections.Generic;

public class GFG {

  static int[] findPeakGrid(int[][] mat)
  {

    // Starting point & end point of Search Space
    int stcol = 0, endcol = mat[0].Length - 1;

    // Bin Search Condition
    while (stcol <= endcol) {

      int midcol = stcol + (endcol - stcol) / 2,
      ansrow = 0;
      // "ansrow" To keep the row number of global
      // Peak element of a column

      // Finding the row number of Global Peak element
      // in Mid Column.
      for (int r = 0; r < mat.Length; r++) {
        ansrow
          = mat[r][midcol] >= mat[ansrow][midcol]
          ? r
          : ansrow;
      }

      // Finding next Search space will be left or
      // right
      bool valid_left = midcol - 1 >= stcol
        && mat[ansrow][midcol - 1]
        > mat[ansrow][midcol];
      bool valid_right = midcol + 1 <= endcol
        && mat[ansrow][midcol + 1]
        > mat[ansrow][midcol];

      // if we're at Peak Element
      if (!valid_left && !valid_right) {
        return new int[] { ansrow, midcol };
      }

      else if (valid_right)
        stcol
        = midcol
        + 1; // move the search space in right
      else
        endcol
        = midcol
        - 1; // move the search space in left
    }

    return new int[] { -1, -1 };
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int[][] arr = { new[] { 9, 8 }, new[] { 2, 6 } };

    int[] result = findPeakGrid(arr);

    Console.WriteLine("Peak element found at index: "
                      + result[0] + "," + result[1]);
  }
}

// This code is contributed by phasing17

Javascript

// Finding peak element in a 2D Array.

function findPeakGrid(mat)
{

    let stcol = 0,  endcol = mat[0].length - 1; // Starting po  end po of Search Space

    while (stcol <= endcol) { // Bin Search Condition

        let midcol = stcol + Math.floor((endcol - stcol) / 2), ansrow = 0;
        // "ansrow" To keep the row number of global Peak
        // element of a column

        // Finding the row number of Global Peak element in
        // Mid Column.
        for (let r = 0; r < mat.length; r++) {
            ansrow = mat[r][midcol] >= mat[ansrow][midcol] ? r : ansrow;
        }

        // Finding next Search space will be left or right
        let valid_left =  midcol - 1 >= stcol && mat[ansrow][midcol - 1] > mat[ansrow][midcol];
        let valid_right = midcol + 1 <= endcol && mat[ansrow][midcol + 1] > mat[ansrow][midcol];

        // if we're at Peak Element
        if (!valid_left && !valid_right) {
            return [ ansrow, midcol ];
        }

        else if (valid_right)
            stcol = midcol  + 1; // move the search space in right
        else
            endcol = midcol  - 1; // move the search space in left
    }

    return [ -1, -1 ];
}


// Driver Code
let arr = [[9, 8], [2 ,6]];

let result = findPeakGrid(arr);

console.log("Peak element found at index: " + result[0] + "," + result[1])

// This code is contributed by phasing14.

Output

Peak element found at index: 0,0

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 

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