Given an array of integers which is initially increasing and then decreasing, find the maximum value in the array.
Examples :
Input: arr[] = {8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1}
Output: 500
Input: arr[] = {1, 3, 50, 10, 9, 7, 6}
Output: 50
Corner case (No decreasing part)
Input: arr[] = {10, 20, 30, 40, 50}
Output: 50
Corner case (No increasing part)
Input: arr[] = {120, 100, 80, 20, 0}
Output: 120
Method 1 (Linear Search): We can traverse the array and keep track of maximum and element. And finally return the maximum element.
Implementation:
C++
// C++ program to find maximum // element #include <bits/stdc++.h>using namespace std;// function to find the maximum element int findMaximum(int arr[], int low, int high) { int max = arr[low]; int i; for (i = low + 1; i <= high; i++) { if (arr[i] > max) max = arr[i]; // break when once an element is smaller than // the max then it will go on decreasing // and no need to check after that else break; } return max; } /* Driver code*/int main() { int arr[] = {1, 30, 40, 50, 60, 70, 23, 20}; int n = sizeof(arr)/sizeof(arr[0]); cout << "The maximum element is " << findMaximum(arr, 0, n-1); return 0; } // This is code is contributed by rathbhupendra |
C
// C program to find maximum// element#include <stdio.h>// function to find the maximum elementint findMaximum(int arr[], int low, int high){int max = arr[low];int i;for (i = low+1; i <= high; i++){ if (arr[i] > max) max = arr[i];// break when once an element is smaller than // the max then it will go on decreasing // and no need to check after that else break;}return max;}/* Driver program to check above functions */int main(){int arr[] = {1, 30, 40, 50, 60, 70, 23, 20};int n = sizeof(arr)/sizeof(arr[0]);printf("The maximum element is %d", findMaximum(arr, 0, n-1));getchar();return 0;} |
Java
// java program to find maximum// elementclass Main{ // function to find the // maximum element static int findMaximum(int arr[], int low, int high) { int max = arr[low]; int i; for (i = low; i <= high; i++) { if (arr[i] > max) max = arr[i]; } return max; } // main function public static void main (String[] args) { int arr[] = {1, 30, 40, 50, 60, 70, 23, 20}; int n = arr.length; System.out.println("The maximum element is "+ findMaximum(arr, 0, n-1)); }} |
Python3
# Python3 program to find # maximum elementdef findMaximum(arr, low, high): max = arr[low] i = low for i in range(high+1): if arr[i] > max: max = arr[i] return max# Driver program to check above functions */arr = [1, 30, 40, 50, 60, 70, 23, 20]n = len(arr)print ("The maximum element is %d"% findMaximum(arr, 0, n-1))# This code is contributed by Shreyanshi Arun. |
C#
// C# program to find maximum// elementusing System;class GFG{ // function to find the // maximum element static int findMaximum(int []arr, int low, int high) { int max = arr[low]; int i; for (i = low; i <= high; i++) { if (arr[i] > max) max = arr[i]; } return max; } // Driver code public static void Main () { int []arr = {1, 30, 40, 50, 60, 70, 23, 20}; int n = arr.Length; Console.Write("The maximum element is "+ findMaximum(arr, 0, n-1)); }}// This code is contributed by Sam007 |
Javascript
<script>// Javascript program to find maximum// element// function to find the maximum elementfunction findMaximum(arr, low, high){ var max = arr[low]; var i; for (i = low + 1; i <= high; i++) { if (arr[i] > max) max = arr[i]; // break when once an element is smaller than // the max then it will go on decreasing // and no need to check after that else break; } return max;}/* Driver code*/var arr = [1, 30, 40, 50, 60, 70, 23, 20];var n = arr.length;document.write("The maximum element is " + findMaximum(arr, 0, n-1));</script> |
PHP
<?php// PHP program to Find the maximum // element in an array which is first// increasing and then decreasingfunction findMaximum($arr, $low, $high){$max = $arr[$low];$i;for ($i = $low; $i <= $high; $i++){ if ($arr[$i] > $max) $max = $arr[$i];}return $max;}// Driver Code$arr = array(1, 30, 40, 50, 60, 70, 23, 20);$n = count($arr);echo "The maximum element is ", findMaximum($arr, 0, $n-1);// This code is contributed by anuj_67.?> |
Output
The maximum element is 70
Time Complexity : O(n)
Auxiliary Space: O(1)
Method 2 (Binary Search – Recursive Solution)
The iterative approach of Binary search to find the maximum element in an array which is first increasing and then decreasing.
The standard binary search approach can be modified in the following ways :-
- If the mid element is greater than both of its adjacent elements, then mid is the maximum.
- If the mid element is smaller than its next element then we should try to search on the right half of the array. So, make, low = mid + 1. Example array : {2, 4, 6, 8, 10, 3, 1}
- If the mid element is greater than the next element, similarly we should try to search on the left half. So, make, high = mid – 1. Example array: {3, 50, 10, 9, 7, 6}
Implementation:
C++
#include <bits/stdc++.h>using namespace std;// Recursive solution for finding bitonic pointint findMaximum(int arr[], int low, int high){ // if there is only one element if (low == high) return arr[high]; // finding the mid index int mid = low + (high - low) / 2; // if the value of mid index is greater than left and right value if (arr[mid] > arr[mid - 1] and arr[mid] > arr[mid + 1]) return arr[mid]; // if value on mid index is greater than next index // shift the search space to right half else if (arr[mid] < arr[mid + 1]) return findMaximum(arr, mid + 1, high); // else vice-versa i.e. to left half else return findMaximum(arr, low, mid - 1);}/* Driver code */int main(){ int arr[] = { 1, 3, 50, 10, 9, 7, 6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "The maximum element is " << findMaximum(arr, 0, n - 1); return 0;}// This is code is contributed by rajdeep999 |
C
#include <stdio.h>int findMaximum(int arr[], int low, int high){ /* Base Case: Only one element is present in arr[low..high]*/ if (low == high) return arr[low]; /* If there are two elements and first is greater than the first element is maximum */ if ((high == low + 1) && arr[low] >= arr[high]) return arr[low]; /* If there are two elements and second is greater than the second element is maximum */ if ((high == low + 1) && arr[low] < arr[high]) return arr[high]; int mid = (low + high)/2; /*low + (high - low)/2;*/ /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1]) return arr[mid]; /* If arr[mid] is greater than the next element and smaller than the previous element then maximum lies on left side of mid */ if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1]) return findMaximum(arr, low, mid-1); else // when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1] return findMaximum(arr, mid + 1, high);}/* Driver program to check above functions */int main(){ int arr[] = {1, 3, 50, 10, 9, 7, 6}; int n = sizeof(arr)/sizeof(arr[0]); printf("The maximum element is %d", findMaximum(arr, 0, n-1)); getchar(); return 0;} |
Java
// java program to find maximum// elementclass Main{ // function to find the // maximum element static int findMaximum(int arr[], int low, int high) { /* Base Case: Only one element is present in arr[low..high]*/ if (low == high) return arr[low]; /* If there are two elements and first is greater than the first element is maximum */ if ((high == low + 1) && arr[low] >= arr[high]) return arr[low]; /* If there are two elements and second is greater than the second element is maximum */ if ((high == low + 1) && arr[low] < arr[high]) return arr[high]; /*low + (high - low)/2;*/ int mid = (low + high)/2; /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1]) return arr[mid]; /* If arr[mid] is greater than the next element and smaller than the previous element then maximum lies on left side of mid */ if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1]) return findMaximum(arr, low, mid-1); else return findMaximum(arr, mid + 1, high); } // main function public static void main (String[] args) { int arr[] = {1, 3, 50, 10, 9, 7, 6}; int n = arr.length; System.out.println("The maximum element is "+ findMaximum(arr, 0, n-1)); }} |
Python3
def findMaximum(arr, low, high): # Base Case: Only one element is present in arr[low..high]*/ if low == high: return arr[low] # If there are two elements and first is greater than # the first element is maximum */ if high == low + 1 and arr[low] >= arr[high]: return arr[low]; # If there are two elements and second is greater than # the second element is maximum */ if high == low + 1 and arr[low] < arr[high]: return arr[high] mid = (low + high)//2 #low + (high - low)/2;*/ # If we reach a point where arr[mid] is greater than both of # its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] # is the maximum element*/ if arr[mid] > arr[mid + 1] and arr[mid] > arr[mid - 1]: return arr[mid] # If arr[mid] is greater than the next element and smaller than the previous # element then maximum lies on left side of mid */ if arr[mid] > arr[mid + 1] and arr[mid] < arr[mid - 1]: return findMaximum(arr, low, mid-1) else: # when arr[mid] is greater than arr[mid-1] and smaller than arr[mid+1] return findMaximum(arr, mid + 1, high) # Driver program to check above functions */arr = [1, 3, 50, 10, 9, 7, 6]n = len(arr)print ("The maximum element is %d"% findMaximum(arr, 0, n-1))# This code is contributed by Shreyanshi Arun. |
C#
// C# program to find maximum// elementusing System;class GFG{ // function to find the // maximum element static int findMaximum(int []arr, int low, int high) { /* Base Case: Only one element is present in arr[low..high]*/ if (low == high) return arr[low]; /* If there are two elements and first is greater than the first element is maximum */ if ((high == low + 1) && arr[low] >= arr[high]) return arr[low]; /* If there are two elements and second is greater than the second element is maximum */ if ((high == low + 1) && arr[low] < arr[high]) return arr[high]; /*low + (high - low)/2;*/ int mid = (low + high)/2; /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1]) return arr[mid]; /* If arr[mid] is greater than the next element and smaller than the previous element then maximum lies on left side of mid */ if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1]) return findMaximum(arr, low, mid-1); else return findMaximum(arr, mid + 1, high); } // main function public static void Main() { int []arr = {1, 3, 50, 10, 9, 7, 6}; int n = arr.Length; Console.Write("The maximum element is "+ findMaximum(arr, 0, n-1)); }}// This code is contributed by Sam007 |
Javascript
<script> function findMaximum( arr, low, high) { /* Base Case: Only one element is present in arr[low..high]*/ if (low == high) return arr[low]; /* If there are two elements and first is greater than the first element is maximum */ if ((high == low + 1) && arr[low] >= arr[high]) return arr[low]; /* If there are two elements and second is greater than the second element is maximum */ if ((high == low + 1) && arr[low] < arr[high]) return arr[high]; mid = (low + high)/2; /*low + (high - low)/2;*/ /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if ( arr[mid] > arr[mid + 1] && arr[mid] > arr[mid - 1]) return arr[mid]; /* If arr[mid] is greater than the next element and smaller than the previous element then maximum lies on left side of mid */ if (arr[mid] > arr[mid + 1] && arr[mid] < arr[mid - 1]) return findMaximum(arr, low, mid-1); // when arr[mid] is greater than arr[mid-1] // and smaller than arr[mid+1] return findMaximum(arr, mid + 1, high); } /* Driver code */ arr = new Array(1, 3, 50, 10, 9, 7, 6); n = arr.length; document.write("The maximum element is" + "\n" + findMaximum(arr, 0, n-1)); // This code is contributed by simranarora5sos </script> |
PHP
<?php// PHP program to Find the maximum // element in an array which is // first increasing and then decreasingfunction findMaximum($arr, $low, $high){ /* Base Case: Only one element is present in arr[low..high]*/ if ($low == $high) return $arr[$low]; /* If there are two elements and first is greater than the first element is maximum */ if (($high == $low + 1) && $arr[$low] >= $arr[$high]) return $arr[$low]; /* If there are two elements and second is greater than the second element is maximum */ if (($high == $low + 1) && $arr[$low] < $arr[$high]) return $arr[$high]; /*low + (high - low)/2;*/ $mid = ($low + $high) / 2; /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element */ if ( $arr[$mid] > $arr[$mid + 1] && $arr[$mid] > $arr[$mid - 1]) return $arr[$mid]; /* If arr[mid] is greater than the next element and smaller than the previous element then maximum lies on left side of mid */ if ($arr[$mid] > $arr[$mid + 1] && $arr[$mid] < $arr[$mid - 1]) return findMaximum($arr, $low, $mid - 1); // when arr[mid] is greater than // arr[mid-1] and smaller than // arr[mid+1] else return findMaximum($arr, $mid + 1, $high);}// Driver Code$arr = array(1, 3, 50, 10, 9, 7, 6);$n = sizeof($arr);echo("The maximum element is "); echo(findMaximum($arr, 0, $n-1));// This code is contributed by nitin mittal.?> |
Output
The maximum element is 50
Time Complexity : O(logn)
Auxiliary Space : O(logn)
This method works only for distinct numbers. For example, it will not work for an array like {0, 1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 5, 3, 3, 2, 2, 1, 1}.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Method 3 (Binary Search – Iterative Solution)
The iterative approach of Binary search to find the maximum element in an array which is first increasing and then decreasing.
The standard binary search approach can be modified in the following ways :-
- If the mid element is greater than both of its adjacent elements, then mid is the maximum.
- If the mid element is smaller than its next element then we should try to search on the right half of the array. So, make, low = mid + 1 .Example array : {2, 4, 6, 8, 10, 3, 1}
- If the mid element is greater than the next element, similarly we should try to search on the left half. So, make, high = mid – 1. Example array : {3, 50, 10, 9, 7, 6}
Implementation:
C++
#include <iostream>using namespace std;int maxInBitonic(int arr[], int low, int high){ // find out the size of the array // for edge case checking int n = high + 1; // main code goes as follows while (low <= high) { // find out the mid int mid = low + (high - low) / 2; // if mid index value is maximum if(arr[mid] > arr[mid+1] and arr[mid] > arr[mid-1]) return arr[mid]; // reducing search space by moving to right else if (arr[mid] < arr[mid + 1]) low = mid + 1; // reducing search space by moving to left else high = mid - 1; } return arr[high];}// Driver functionint main(){ int arr[] = { 1, 3, 50, 10, 9, 7, 6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "The maximum element is " << maxInBitonic(arr, 0, n - 1); return 0;}// This code is contributed by rajdeep999 |
Java
import java.util.*;class GFG{static int maxInBitonic(int arr[], int l, int r){ while (l <= r) { int m = l + (r - l) / 2; // m = (l + r) / 2 /****Base Cases Starts*****/ if(l==r) return arr[l]; /* If there are two elements and first is greater then the first element is maximum */ if ((r == l + 1) && arr[l] >= arr[r]) return arr[l]; /* If there are two elements and second is greater then the second element is maximum */ if ((r == l + 1) && arr[l] < arr[r]) return arr[r]; /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1]) return arr[m]; /****Base Case ends *****/ // move to left with l and r=m-1 if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1]) r = m - 1; else l = m + 1; // move to right with l=m+1 and r } // if we reach here, then element was // not present return -1;}// Driver functionpublic static void main(String[] args){ int arr[] = { 1, 3, 50, 10, 9, 7, 6 }; int n = arr.length; System.out.print("The maximum element is " + maxInBitonic(arr, 0, n - 1));}}// This code is contributed by todaysgaurav |
Python3
# Python 3 program for the above approachdef maxInBitonic(arr, l, r) : while (l <= r) : m = int(l + (r - l) / 2) # m = (l + r) / 2 #Base Cases Starts*****/ if(l==r) return arr[l]; # If there are two elements and first is greater # then the first element is maximum */ if ((r == l + 1) and arr[l] >= arr[r]): return arr[l] # If there are two elements and second is greater # then the second element is maximum */ if ((r == l + 1) and arr[l] < arr[r]): return arr[r] # If we reach a point where arr[mid] is greater # than both of its adjacent elements arr[mid-1] and # arr[mid+1], then arr[mid] is the maximum # element*/ if (arr[m] > arr[m + 1] and arr[m] > arr[m - 1]): return arr[m] #***Base Case ends *****/ # move to left with l and r=m-1 if (arr[m] > arr[m + 1] and arr[m] < arr[m - 1]) : r = m - 1 else : l = m + 1 # move to right with l=m+1 and r # if we reach here, then element was # not present return -1# Driver functionarr = [ 1, 3, 50, 10, 9, 7, 6 ]n = len(arr)print("The maximum element is ", maxInBitonic(arr, 0, n - 1))# This code is contributed by splevel62. |
C#
using System;class GFG{static int maxInBitonic(int []arr, int l, int r){ while (l <= r) { int m = l + (r - l) / 2; // m = (l + r) / 2 /****Base Cases Starts*****/ if(l==r) return arr[l]; /* If there are two elements and first is greater then the first element is maximum */ if ((r == l + 1) && arr[l] >= arr[r]) return arr[l]; /* If there are two elements and second is greater then the second element is maximum */ if ((r == l + 1) && arr[l] < arr[r]) return arr[r]; /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1]) return arr[m]; /****Base Case ends *****/ // move to left with l and r=m-1 if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1]) r = m - 1; else l = m + 1; // move to right with l=m+1 and r } // if we reach here, then element was // not present return -1;}// Driver functionpublic static void Main(String[] args){ int []arr = { 1, 3, 50, 10, 9, 7, 6 }; int n = arr.Length; Console.Write("The maximum element is " + maxInBitonic(arr, 0, n - 1));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// JavaScript programfunction maxInBitonic(arr, l, r){ while (l <= r) { var m = l + (r - l) / 2; // m = (l + r) / 2 /****Base Cases Starts*****/ if(l==r) return arr[l]; /* If there are two elements and first is greater then the first element is maximum */ if ((r == l + 1) && arr[l] >= arr[r]) return arr[l]; /* If there are two elements and second is greater then the second element is maximum */ if ((r == l + 1) && arr[l] < arr[r]) return arr[r]; /* If we reach a point where arr[mid] is greater than both of its adjacent elements arr[mid-1] and arr[mid+1], then arr[mid] is the maximum element*/ if (arr[m] > arr[m + 1] && arr[m] > arr[m - 1]) return arr[m]; /****Base Case ends *****/ // move to left with l and r=m-1 if (arr[m] > arr[m + 1] && arr[m] < arr[m - 1]) r = m - 1; else l = m + 1; // move to right with l=m+1 and r } // if we reach here, then element was // not present return -1;}// Driver function var arr = [ 1, 3, 50, 10, 9, 7, 6 ]; var n = arr.length; document.write("The maximum element is " + maxInBitonic(arr, 0, n - 1));// This code is contributed by shivanisinghss2110 </script> |
Output
The maximum element is 50
Time Complexity: O(log n)
Auxiliary Space: O(1)
Method 4 (Using Stack) :
1.Create an empty stack to hold the indices of the array elements.
2.Traverse the array from left to right until we find the maximum element. Push the index of each element onto the
stack as long as the element is less than or equal to the previous element.
3.Once we find an element that is greater than the previous element, we know that the maximum element has been
reached. We can then pop all the indices from the 4.stack until we find an index whose corresponding element
is greater than the current element.
4.The maximum element is the element corresponding to the last index remaining on the stack.
Implementation of above approach :
C++
#include <bits/stdc++.h>using namespace std;int findMax(int arr[], int n){ stack<int> s; int max = 0; // traverse the array from left to right for (int i = 0; i < n; i++) { // push the index onto the stack if the element is // less than or equal to the previous element if (s.empty() || arr[i] <= arr[s.top()]) { s.push(i); } else { // pop all the indices from the stack until we // find an index whose corresponding element is // greater than the current element while (!s.empty() && arr[i] > arr[s.top()]) { int index = s.top(); s.pop(); // update the maximum element if (arr[index] > max) { max = arr[index]; } } // push the current index onto the stack s.push(i); } } // the maximum element is the element corresponding to // the last index remaining on the stack while (!s.empty()) { int index = s.top(); s.pop(); if (arr[index] > max) { max = arr[index]; } } return max;}int main(){ int arr[] = { 1, 3, 50, 10, 9, 7, 6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << "The maximum element is " << findMax(arr, n); return 0;} |
Java
import java.util.Stack;public class Main { public static int findMax(int[] arr, int n) { Stack<Integer> s = new Stack<>(); int max = 0; // traverse the array from left to right for (int i = 0; i < n; i++) { // push the index onto the stack if the element // is less than or equal to the previous element if (s.empty() || arr[i] <= arr[s.peek()]) { s.push(i); } else { // pop all the indices from the stack until // we find an index whose corresponding // element is greater than the current // element while (!s.empty() && arr[i] > arr[s.peek()]) { int index = s.peek(); s.pop(); // update the maximum element if (arr[index] > max) { max = arr[index]; } } // push the current index onto the stack s.push(i); } } // the maximum element is the element corresponding // to the last index remaining on the stack while (!s.empty()) { int index = s.peek(); s.pop(); if (arr[index] > max) { max = arr[index]; } } return max; } public static void main(String[] args) { int[] arr = { 1, 3, 50, 10, 9, 7, 6 }; int n = arr.length; System.out.println("The maximum element is " + findMax(arr, n)); }} |
Python3
def findMax(arr, n): s = [] max = 0 # traverse the array from left to right for i in range(n): # push the index onto the stack if the element is # less than or equal to the previous element if not s or arr[i] <= arr[s[-1]]: s.append(i) else: # pop all the indices from the stack until we # find an index whose corresponding element is # greater than the current element while s and arr[i] > arr[s[-1]]: index = s.pop() # update the maximum element if arr[index] > max: max = arr[index] # push the current index onto the stack s.append(i) # the maximum element is the element corresponding to # the last index remaining on the stack while s: index = s.pop() if arr[index] > max: max = arr[index] return maxarr = [1, 3, 50, 10, 9, 7, 6]n = len(arr)print("The maximum element is", findMax(arr, n)) |
C#
// C# program to find maximum elementusing System;using System.Collections.Generic;class GFG{ static int FindMax(int[] arr, int n) { Stack<int> stack = new Stack<int>(); int max = 0; // Traverse the array from left to right for (int i = 0; i < n; i++) { // Push the index onto the stack if the element is // less than or equal to the previous element if (stack.Count == 0 || arr[i] <= arr[stack.Peek()]) { stack.Push(i); } else { // Pop all the indices from the stack until we // find an index whose corresponding element is // greater than the current element while (stack.Count > 0 && arr[i] > arr[stack.Peek()]) { int index = stack.Pop(); // Update the maximum element if (arr[index] > max) { max = arr[index]; } } // Push the current index onto the stack stack.Push(i); } } // The maximum element is the element corresponding to // the last index remaining on the stack while (stack.Count > 0) { int index = stack.Pop(); if (arr[index] > max) { max = arr[index]; } } return max; }//Driver Code static void Main() { int[] arr = { 1, 3, 50, 10, 9, 7, 6 }; int n = arr.Length; Console.WriteLine("The maximum element is " + FindMax(arr, n)); }} |
Javascript
function findMax(arr) { const stack = []; // Create an array to simulate the stack let max = 0; // Variable to store the maximum element found // Traverse the array from left to right for (let i = 0; i < arr.length; i++) { // Push the index onto the stack if the element is less than or equal to the previous element if (stack.length === 0 || arr[i] <= arr[stack[stack.length - 1]]) { stack.push(i); } else { // Pop all the indices from the stack until we find an index whose corresponding element is greater than the current element while (stack.length > 0 && arr[i] > arr[stack[stack.length - 1]]) { const index = stack.pop(); // Update the maximum element if the element at the popped index is greater than the current max if (arr[index] > max) { max = arr[index]; } } // Push the current index onto the stack stack.push(i); } } // After processing all elements, there may be remaining indices on the stack. // These indices correspond to elements that are greater than any elements to their right. // We need to check these elements and update the max value if necessary. while (stack.length > 0) { const index = stack.pop(); if (arr[index] > max) { max = arr[index]; } } return max;}// Driver codeconst arr = [1, 3, 50, 10, 9, 7, 6];console.log("The maximum element is " + findMax(arr)); |
Output
The maximum element is 50
Time Complexity : O(n)
Auxiliary Space : O(n)
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