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++
// 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]; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
PHP
<?php // PHP program to find the smallest number // in an array of decrease and increasing numbers // Function to find the smallest // number's index function minimal($a, $n) { $lo = 0; $hi = $n - 1; // Do a binary search while ($lo < $hi) { // Find the mid element $mid = ($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 $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)
