Last Updated on March 28, 2022 by Ria Pathak
CONCEPTS USED:
Binary Search
DIFFICULTY LEVEL:
Medium
PROBLEM STATEMENT(SIMPLIFIED):
Given an array
AofNintegers, such that till a point these integers are strictly increasing and after that strictly decreasing, find the element present at that point.
See original problem statement here
For Example:
Input : A = [10, 12, 14, 15, 8, 7, 6]
Output : 15
Explanation : 15 is the top point. As before 15 all elements are strictly increasing and after 15 all elements are strictly decreasing.Can we use Binary Search here ?
Given that half of the array is sorted in increasing order and the other half in decreasing order and we need to find an element that is greater than both elements, one on the left and one on the right. So
Binary Searchcould be an efficient alternative to quickly search for that element inLogarithmic Time Complexity.
OBSERVATION:
The desired element would be the
Maximumelement of the array.
SOLVING APPROACH:
The idea is to use
Binary Search.Initialize
startandendas starting and ending indexes of the array.Calculate,
mid = start + (end - start) / 2.For every such
midcalculated check if the element atmidis greater than both of its left and right element. IfYesreturn this element.Else if element at
midis only greater than left element, then go on searching in the right half.Else, go on searching in the left half till the element is found.
ILLUSTRATION:
A = [12, 14, 15, 8, 7, 6, 5]
i = 0
j = 6
mid = (0 + 6) / 2 = 3
Since, A[3] < A[2]
j = mid - 1 = 2
i = 0
j = 2
mid = (0 + 2) / 2 = 1
Since, A[1] < A[2]
i = mid + 1 = 2
i = 2
j = 2
mid = (2 + 2) / 2 = 2
Since, A[2] > A[1] and A[2] > A[3]
Therefore, A[2] = 15 is our required top point. SOLUTIONS:
#include <stdio.h>
/* function for find Mountain Top */
int MountainTop(int *arr, int start, int end){
while(start <= end){
int mid = start + (end - start) / 2;
/*if mid index element is greater than both of
its adjacent elements, this is our top element*/
if(arr[mid] > arr[mid - 1] && arr[mid] > arr[mid + 1])
return arr[mid];
/* if mid indexed element is only greater than left element
this implies top is in the right half */
else if(arr[mid] > arr[mid - 1])
start = mid + 1;
/* if mid indexed element is only greater than right element
this implies top is in the left half */
else
end = mid - 1;
}
}
int main()
{
int t; scanf("%d", &t);
while(t--){
int n; scanf("%d", &n);
int arr[n];
for(int i=0; i<n; i++)
scanf("%d", &arr[i]);
printf("%d\n", MountainTop(arr, 0, n-1));
}
return 0;
}
#include <bits/stdc++.h>
using namespace std;
/* function for find Mountain Top */
int MountainTop(int *arr, int start, int end){
while(start <= end){
int mid = start + (end - start) / 2;
/*if mid index element is greater than both of
its adjacent elements, this is our top element*/
if(arr[mid] > arr[mid - 1] && arr[mid] > arr[mid + 1])
return arr[mid];
/* if mid indexed element is only greater than left element
this implies top is in the right half */
else if(arr[mid] > arr[mid - 1])
start = mid + 1;
/* if mid indexed element is only greater than right element
this implies top is in the left half */
else
end = mid - 1;
}
}
int main()
{
int t; cin>>t;
while(t--){
int n; cin>>n;
int arr[n];
for(int i=0; i<n; i++)
cin>>arr[i];
cout<<MountainTop(arr, 0, n-1)<<"\n";
}
return 0;
}
import java.util.*;
import java.io.*;
public class Main {
/* function for find Mountain Top */
static int MountainTop(int []arr, int start, int end){
while(start <= end){
int mid = start + (end - start) / 2;
/*if mid index element is greater than both of
its adjacent elements, this is our top element*/
if(arr[mid] > arr[mid - 1] && arr[mid] > arr[mid + 1])
return arr[mid];
/* if mid indexed element is only greater than left element
this implies top is in the right half */
else if(arr[mid] > arr[mid - 1])
start = mid + 1;
/* if mid indexed element is only greater than right element
this implies top is in the left half */
else
end = mid - 1;
}
// if no mountain top is present
return 0;
}
public static void main(String args[]) throws IOException {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t != 0){
int n = sc.nextInt();
int arr[] = new int[n];
for(int i=0; i<n; i++)
arr[i] = sc.nextInt();
System.out.println(MountainTop(arr, 0, n-1));
t--;
}
}
}
Space Complexity: O(1)
[forminator_quiz id="1528"]
This article tried to discuss Binary Search. Hope this blog helps you understand and solve the problem. To practice more problems on Searching you can check out MYCODE | Competitive Programming.