Minimum in an array which is first decreasing then increasing

Given an array of N integers where array elements form a strictly decreasing and increasing sequence. The task is to find the smallest number in such an array.

Constraints: N >= 3

Examples:

Input: a[] = {2, 1, 2, 3, 4}
Output: 1

Input: a[] = {8, 5, 4, 3, 4, 10}
Output: 3 


A naive approach is to linearly traverse the array and find out the smallest number. The time complexity will be thus O(N).

An efficient approach is to modify the binary search and use it. Divide the array into two halves use binary search to check if a[mid] < a[mid+1] or not. If a[mid] < a[mid+1], then the smallest number lies in the first half which is low to mid, else it lies in the second half which is mid+1 to high.

Algorithm:

while(lo > 1

    if a[mid] < a[mid+1] then hi = mid

    else lo = mid+1
} 

Below is the implementation of the above approach:

C++

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// C++ program to find the smallest number
// in an array of decrease and increasing numbers
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the smallest number's index
int minimal(int a[], int n)
{
    int lo = 0, hi = n - 1;
  
    // Do a binary search
    while (lo < hi) {
  
        // Find the mid element
        int mid = (lo + hi) >> 1;
  
        // Check for break point
        if (a[mid] < a[mid + 1]) {
            hi = mid;
        }
        else {
            lo = mid + 1;
        }
    }
  
    // Return the index
    return lo;
}
  
// Driver Code
int main()
{
    int a[] = { 8, 5, 4, 3, 4, 10 };
    int n = sizeof(a) / sizeof(a[0]);
    int ind = minimal(a, n);
  
    // Print the smallest number
    cout << a[ind];
}

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Java

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// Java program to find the smallest number 
// in an array of decrease and increasing numbers 
class Solution
{
// Function to find the smallest number's index 
static int minimal(int a[], int n) 
    int lo = 0, hi = n - 1
  
    // Do a binary search 
    while (lo < hi) { 
  
        // Find the mid element 
        int mid = (lo + hi) >> 1
  
        // Check for break point 
        if (a[mid] < a[mid + 1]) { 
            hi = mid; 
        
        else
            lo = mid + 1
        
    
  
    // Return the index 
    return lo; 
  
// Driver Code 
public static void main(String args[]) 
    int a[] = { 8, 5, 4, 3, 4, 10 }; 
    int n = a.length; 
    int ind = minimal(a, n); 
  
    // Print the smallest number 
    System.out.println( a[ind]); 
}
//contributed by Arnab Kundu

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Python3

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# Python 3 program to find the smallest 
# number in a array of decrease and
# increasing numbers
  
# function to find the smallest 
# number's index
def minimal(a, n):
      
    lo, hi = 0, n - 1
      
    # Do a binary search
    while lo < hi:
          
        # find the mid element
        mid = (lo + hi) // 2
          
        # Check for break point
        if a[mid] < a[mid + 1]:
            hi = mid
        else:
            lo = mid + 1
        return lo 
      
# Driver code
a = [8, 5, 4, 3, 4, 10]
  
n = len(a)
  
ind = minimal(a, n)
  
# print the smallest number
print(a[ind])
  
# This code is contributed
# by Mohit Kumar

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C#

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// C# program to find the smallest number 
// in an array of decrease and increasing numbers 
using System;
class Solution
{
// Function to find the smallest number's index 
static int minimal(int[] a, int n) 
    int lo = 0, hi = n - 1; 
  
    // Do a binary search 
    while (lo < hi) { 
  
        // Find the mid element 
        int mid = (lo + hi) >> 1; 
  
        // Check for break point 
        if (a[mid] < a[mid + 1]) { 
            hi = mid; 
        
        else
            lo = mid + 1; 
        
    
  
    // Return the index 
    return lo; 
  
// Driver Code 
public static void Main() 
    int[] a = { 8, 5, 4, 3, 4, 10 }; 
    int n = a.Length; 
    int ind = minimal(a, n); 
  
    // Print the smallest number 
    Console.WriteLine( a[ind]); 
}
//contributed by Mukul singh

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PHP

> 1;

// Check for break point
if ($a[$mid] < $a[$mid + 1]) { $hi = $mid; } else { $lo = $mid + 1; } } // Return the index return $lo; } // Driver Code $a = array( 8, 5, 4, 3, 4, 10 ); $n = sizeof($a); $ind = minimal($a, $n); // Print the smallest number echo $a[$ind]; // This code is contributed // by Sach_Code ?>

Output:

3


Time Complexity
: O(Log N)
Auxiliary Space: O(1)



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Striver(underscore)79 at Codechef and codeforces D

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