Find the Peak Element in a 2D Array/Matrix
Last Updated :
11 Sep, 2024
Given a 2D Array/Matrix mat[][], 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 peak element is not necessarily the overall maximal element. It only needs to be greater than existing adjacent
- More than one such element can exist. We need to return any of them
- 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
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 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)
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));
}
}
# 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# 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)
// 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)
OutputPeak element found at index: 0, 0
Time Complexity: O(rows * columns)
Auxiliary Space: O(1)
Efficient Approach to Find Peak Element in Matrix
This problem is mainly an extension of Find a peak element in 1D array. We apply similar Binary Search based solution here, as shown below:
- 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 >= mat[index][mid-1] & max >= mat[index][mid+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 implementation of the above algorithm:
C++
// Finding peak element in a 2D Array.
#include <bits/stdc++.h>
using namespace std;
vector<int> findPeakGrid(vector<vector<int> >& mat)
{
// Starting point & end point of Search Space
int stcol = 0, endcol = mat[0].size() - 1;
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;
}
// 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
# 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.
// 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
// 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.
OutputPeak 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
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