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5/12/12 Find the maximum element in an array which is first increasing and then decreasing | GeeksforGeeks

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Find the maximum element in an array which is firstincreasing and then decreasing

January 14, 2012

Given an array of integers which is initially increasing and then decreasing, find the maximumvalue in the array.

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 themaximum element.

Time Complexity: O(n)

Method 2 (Binary Search)We can modify the standard Binary Search algorithm for the given type of arrays.i) If the mid element is greater than both of its adjacent elements, then mid is the maximum.ii) If mid element is greater than its next element and smaller than the previous element thenmaximum lies on left side of mid. Example array: {3, 50, 10, 9, 7, 6}iii) If mid element is smaller than its next element and greater than the previous element thenmaximum lies on right side of mid. Example array: {2, 4, 6, 8, 10, 3, 1}

#include <stdio.h> 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 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;}

#include <stdio.h>



Time Complexity: 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}.

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 then 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 then 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;}



Please write comments if you find anything incorrect, or you want to share more informationabout the topic discussed above.

You may also like following posts

1. Search an element in a sorted and pivoted array2. Check for Majority Element in a sorted array3. Given an array arr[], find the maximum j – i such that arr[j] > arr[i]4. Maximum and minimum of an array using minimum number of comparisons5. Find a Fixed Point in a given array

20 comments so far

1. jk says:March 23, 2012 at 10:04 AM

Send 2people

int max(int a, int b){return (a > b) ? a : b;} int maxElementRec(int* array, int startIndex, int endIndex){ if(startIndex == endIndex) return array[startIndex]; if(array[startIndex] > array[startIndex + 1]) return array[startIndex]; if(array[endIndex] > array[endIndex - 1]) return array[endIndex]; int midIndex = (startIndex + endIndex)/2; if(array[midIndex] > array[midIndex - 1] && array[midIndex] > array[midIndex + 1]) if(array[midIndex] > array[midIndex - 1] || array[midIndex] < array[midIndex + 1]) if(array[midIndex] < array[midIndex - 1] || array[midIndex] > array[midIndex + 1]) return max(maxElementRec(array, startIndex, midIndex - 1), maxElementRec(array, midIndex + 1, endIndex));} int maxElement(int* array, int size){ return maxElementRec(array, 0, size - 1);} int main(){ int arr[] = {0, 1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 5, 5, 3, 3, 2, 2, 1, 1}; printf("%d\n", maxElement(arr, 19));



Reply2. kartikaditya says:

February 20, 2012 at 3:39 AM

}

public class MaxInIncAndDec { private static int getMaxInIncAndDec(int a[], int start, int if (start > end) { return Integer.MIN_VALUE; } int mid = (start + end) >> 1; while (start < end) { if (a[mid] > a[mid + 1]) { end = mid; } else if (a[mid] < a[mid + 1]) { start = mid + 1; } else { int leftMax = getMaxInIncAndDec(a, start, mid - 1); int rightMax = getMaxInIncAndDec(a, mid + 1, end); if (leftMax == Integer.MIN_VALUE) { return rightMax; } if (rightMax == Integer.MIN_VALUE) { return leftMax; } return (leftMax > rightMax) ? leftMax : rightMax; } mid = (start + end) >> 1; } return a[mid]; } public static int getMaxInIncAndDec(int a[]) { return getMaxInIncAndDec(a, 0, a.length - 1); } public static void main(String args[]) { System.out.println(getMaxInIncAndDec(new int[]{8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1})); System.out.println(getMaxInIncAndDec(new int[]{1, 3, 50, 10, 9, 7, 6})); System.out.println(getMaxInIncAndDec(new int[]{10, 20, 30, 40, 50})); System.out.println(getMaxInIncAndDec(new int[]{120, 100, 80, 20, 0})); System.out.println(getMaxInIncAndDec(new int[]{0, 1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 5, 3, 3, 2, 2, 1, 1})); }}



Reply3. Krishs says:

January 27, 2012 at 6:36 PM

Well simple enough ...

Replykartik says:January 27, 2012 at 11:49 PM

This looks similar to method 1 (linear search). Time complexity of this would beO(n).

Reply4. Wayne says:

January 24, 2012 at 12:40 AM

If the array elements aren't necessarily distinct, then in the extreme case, all elementsare the same. Method 2 would degrade to O(N) solution since you have to scan bothsides linearly until you meet anything breaking the even.

Reply5. sandeep says:


what if the numbers are repeated.like (0 1 1 2 2 2 2 2 3 4 4 5 3 3 2 2 1 1) and you findmid whose left and right both element are equal to mid how will you decide which partto check.. left or right?

ReplyGeeksforGeeks says:January 18, 2012 at 9:51 PM

int max_ ( int arr[] , int len ) { int index = 0 ; for( index = 0; index < len ; index++ ) { if( index != (len -1) && arr[index] > arr[index+1] ) { } return arr[len-1]; }

/* Paste your code here (You may delete these lines if not writing code) */



@sandeep: Thanks for pointing out this case. The method 2 doesn't always workif there are duplicates in array. We have added the same to the original post.Keep it up!

ReplyShreyas says:January 30, 2012 at 11:27 AM

@Sandeep:they said increasing and then decreasing so no duplicates allowed. 1 2 3 0 -1 butcan never be 1 2 2 3.

Reply6. shaan7 says:

January 17, 2012 at 11:43 AM

Why not start from the beginning and when you find a number less than previousnumber, the previous number is max?

ReplyNithish says:January 17, 2012 at 2:01 PM

@shaan: That would have a worst case complexity of o(N). Consider the examplegiven in the problem itself - Input: arr[] = {10, 20, 30, 40, 50}.

Replyshaan7 says:January 17, 2012 at 2:41 PM

Ah yes, my bad, actually I was suggesting an improvement for Method1.

Replyabbie says:February 17, 2012 at 1:09 PM

@Nithish: since the array is first increasing and then decreasing so weneed not to compare all the elements in Method 1, the moment wefind any element greater than current max element we can stop theiteration in Method 1. It'll reduce the no of comparisons though it'llnot improve the complexity that much.

Correct me if I am wrong. Thanx




Reply7. Priyanka says:

January 17, 2012 at 12:39 AM

Replygowtham says:February 12, 2012 at 10:02 AM

Replygowtham says:February 12, 2012 at 10:06 AM

i am new here so correct me if i am wrong. @priyanka you have used m

8. Ankur says:

#include<stdio.h>#define n 8void main(){int a[n];int l=0,i,h=n-1,m;for(i=0;i<n;i++)scanf("%d",&a[i]);while(l<=h){ m=(l+h)/2; if(m==n-1) break; else if(a[m]>a[m+1] && a[m]>a[m-1] && m<n-1 && m>0) break; else if(a[m]<a[m+1] && m<n-1) l=m+1; else if(a[m]>a[m+1] && m<n-1) h=m-1;}printf("%d",a[m]);}

i am new here so correct me if i am wrong. @priyanka you have used m<n-1 in every

<p></p> <div class="reply"> <a class="comment-reply-link" href="17028?replytocom=7479#respond"



January 16, 2012 at 3:19 AM

Iterative code below which is easier to understand

Reply9. flyinghearts says:


Reply10. sachin says:

January 15, 2012 at 5:24 AM

This will not will for

int findMaximum(int a[], int n){ int low=0; int high=n-1; int mid; while(low<=high){ mid=low+(high-low)/2; if((mid==0 || a[mid-1]<a[mid]) && (mid==n-1 || a[mid]>a[mid+1])){ cout<<a[mid]<<" "; return a[mid]; }/*Increasing Part of Array*/ else if((mid==0 || a[mid-1]<a[mid]) && (mid==n-1 || a[mid]<a[mid+1])){ cout<<a[mid]<<" "; low=mid+1; }/*Decreasing Part of Array */ else if((mid==0 || a[mid-1]>a[mid]) && (mid==n-1 || a[mid]>a[mid+1])) high=mid-1; }}

int findMaximum(const int arr[], size_t len){ assert(len > 0); size_t low = 0, high = len - 1; while (low != high) { size_t mid = low + (high - low) / 2; if (arr[mid] < arr[mid + 1]) low = mid + 1; else high = mid; } return arr[low];}



int[] a = {10,15,20,12,5,6,25};It will return 20 instead of 25

ReplyNithish says:January 15, 2012 at 5:51 AM

@sachin: The question reads - "Given an array of integers which is initiallyincreasing and then decreasing,...". So your input is not for this question.

Replysachin says:January 15, 2012 at 5:55 AM

Ohh I see ! thanks nitish

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