Naive Approach to Find Peak Element in Matrix
Iterate through all the elements of the Matrix and check if it is greater/equal to all its neighbors. 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 Codeint 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 arraydef 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 codearr = [[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 arrayusing 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 arrayfunction 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 codelet 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)
Find the Peak Element in a 2D Array/Matrix
Given a 2D Array/Matrix, the task is to find the Peak element.
An element is a peak element if it is greater than or equal to its four neighbors, left, right, top and bottom.
- A Diagonal adjacent is not considered a neighbour.
- A peak element is not necessarily the maximal element.
- More than one such element can exist.
- There is always a peak element.
- For corner elements, missing neighbors are considered of negative infinite value.
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