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
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.
- 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 >= A[index][mid-1] & max >= A[index][pick+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 above algorithm:
C++
// Finding peak element in a 2D Array. #include <bits/stdc++.h> 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; } |
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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 |
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Python3
# Finding peak element in a 2D Array. MAX = 100from 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 = 4columns = 4print(findPeak(arr, rows, columns)) # This code is contributed by Mohit Kumar |
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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. |
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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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