# Minimum in an array which is first decreasing then increasing

Last Updated : 29 Aug, 2023

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: 1Input: a[] = {8, 5, 4, 3, 4, 10}Output: 3 `

A naive approach is to linearly traverse the array and find out the smallest number.

## C++

 `// CPP program to find minimum element` `// in an array.` `#include ` `using` `namespace` `std;`   `int` `minimal(``int` `arr[], ``int` `n)` `{` `    ``int` `ans = arr[0];` `    ``for` `(``int` `i = 1; i < n; i++)` `        ``ans = min(ans, arr[i]);` `    ``return` `ans;` `}`     `int` `main()` `{` `    ``int` `arr[] = { 8, 5, 4, 3, 4, 10 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << minimal(arr, n);` `    ``return` `0;` `}` `// This code is contributed by Vishal Dhaygude`

## Java

 `import` `java.util.*;`   `public` `class` `Main {` `    `  `    ``public` `static` `int` `minimal(``int``[] arr, ``int` `n) {` `        ``int` `ans = arr[``0``];` `        ``for` `(``int` `i = ``1``; i < n; i++) {` `            ``ans = Math.min(ans, arr[i]);` `        ``}` `        ``return` `ans;` `    ``}` `    `  `    ``public` `static` `void` `main(String[] args) {` `        ``int``[] arr = { ``8``, ``5``, ``4``, ``3``, ``4``, ``10` `};` `        ``int` `n = arr.length;` `        ``System.out.println(minimal(arr, n));` `    ``}` `}`

## Python3

 `#python code` `def` `minimal(arr, n):` `    ``ans ``=` `arr[``0``]` `    ``for` `i ``in` `range``(``1``, n):` `        ``ans ``=` `min``(ans, arr[i])` `    ``return` `ans`   `arr ``=` `[``8``, ``5``, ``4``, ``3``, ``4``, ``10``]` `n ``=` `len``(arr)` `print``(minimal(arr, n))`

## C#

 `// C# program to find minimum element` `// in an array.` `using` `System;`   `public` `class` `Program {` `    ``public` `static` `int` `Minimal(``int``[] arr, ``int` `n)` `    ``{` `        ``int` `ans = arr[0];` `        ``for` `(``int` `i = 1; i < n; i++)` `            ``ans = Math.Min(ans, arr[i]);` `        ``return` `ans;` `    ``}` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 8, 5, 4, 3, 4, 10 };` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(Minimal(arr, n));` `    ``}` `}` `// This code is contributed by user_dtewbxkn77n`

## Javascript

 `// JavaScript program to find minimum element` `// in an array.`   `function` `minimal(arr, n) {` `    ``let ans = arr[0];` `    ``for` `(let i = 1; i < n; i++) {` `        ``ans = Math.min(ans, arr[i]);` `    ``}` `    ``return` `ans;` `}`   `let arr = [8, 5, 4, 3, 4, 10];` `let n = arr.length;` `console.log(minimal(arr, n));`

Output

```3

```

Time Complexity: O(N), we need to use a loop to traverse N times linearly.

Auxiliary Space: O(1), as we are not using any extra space.

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 ` `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`

## Javascript

 ``

## 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), as we are using binary search, in binary search in each traversal we reduce by division of 2 so the effective time will be 1+1/2+1/4+… which is equivalent to logN.
Auxiliary Space: O(1), as we are not using any extra space.
Method 3(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 minimum element. Push the index of each element onto the stack as long as the element is greater than or equal to the previous element.3.Once we find an element that is lesser than the previous element, we know that the minimum element has beenreached. We can then pop all the indices from the 4.stack until we find an index whose corresponding element is less than the current element.4.The minimum element is the element corresponding to the last index remaining on the stack.`

Implementation of the above code :

## C++

 `#include ` `using` `namespace` `std;`   `int` `findMin(``int` `arr[], ``int` `n)` `{` `    ``stack<``int``> s;` `    ``int` `min = 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` `        ``// greater 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` `            ``// less than the current element` `            ``while` `(!s.empty() && arr[i] < arr[s.top()]) {` `                ``int` `index = s.top();` `                ``s.pop();` `                ``// update the minimum element` `                ``if` `(arr[index] < arr[min]) {` `                    ``min = index;` `                ``}` `            ``}` `            ``// push the current index onto the stack` `            ``s.push(i);` `        ``}` `    ``}`   `    ``// the minimum 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] < arr[min]) {` `            ``min = index;` `        ``}` `    ``}`   `    ``return` `arr[min];` `}`   `int` `main()` `{`   `    ``int` `arr[] = { 50, 20, 3, 1, 6, 7, 10, 56 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << ``"The minimum element is "` `<< findMin(arr, n);`   `    ``return` `0;` `}`

## Java

 `import` `java.util.*;`   `public` `class` `Main {` `public` `static` `int` `findMin(``int``[] arr, ``int` `n) {` `Stack s = ``new` `Stack<>();` `int` `min = ``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` `        ``// greater 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` `            ``// less than the current element` `            ``while` `(!s.empty() && arr[i] < arr[s.peek()]) {` `                ``int` `index = s.peek();` `                ``s.pop();` `                ``// update the minimum element` `                ``if` `(arr[index] < arr[min]) {` `                    ``min = index;` `                ``}` `            ``}` `            ``// push the current index onto the stack` `            ``s.push(i);` `        ``}` `    ``}`   `    ``// the minimum 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] < arr[min]) {` `            ``min = index;` `        ``}` `    ``}`   `    ``return` `arr[min];` `}`   `public` `static` `void` `main(String[] args) {` `    ``int``[] arr = { ``50``, ``20``, ``3``, ``1``, ``6``, ``7``, ``10``, ``56` `};` `    ``int` `n = arr.length;` `    ``System.out.println(``"The minimum element is "` `+ findMin(arr, n));` `}` `}`

## Python3

 `# code` `from` `typing ``import` `List` `import` `sys`     `def` `findMin(arr: ``List``[``int``], n: ``int``) ``-``> ``int``:` `    ``s ``=` `[]` `    ``min_index ``=` `0`   `    ``# traverse the array from left to right` `    ``for` `i ``in` `range``(n):` `        ``# push the index onto the stack if the element is` `        ``# greater 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` `            ``# less than the current element` `            ``while` `s ``and` `arr[i] < arr[s[``-``1``]]:` `                ``index ``=` `s.pop()` `                ``# update the minimum element` `                ``if` `arr[index] < arr[min_index]:` `                    ``min_index ``=` `index` `            ``# push the current index onto the stack` `            ``s.append(i)`   `    ``# the minimum element is the element corresponding to` `    ``# the last index remaining on the stack` `    ``while` `s:` `        ``index ``=` `s.pop()` `        ``if` `arr[index] < arr[min_index]:` `            ``min_index ``=` `index`   `    ``return` `arr[min_index]`     `# Driver code` `arr ``=` `[``50``, ``20``, ``3``, ``1``, ``6``, ``7``, ``10``, ``56``]` `n ``=` `len``(arr)` `print``(``"The minimum element is"``, findMin(arr, n))`

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `GFG{` `    `  `    ``static` `int` `findMin(``int``[] arr, ``int` `n)` `    ``{` `        ``Stack<``int``> s = ``new` `Stack<``int``>();` `        ``int` `min = 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` `            ``// greater than or equal to the previous element` `            ``if` `(s.Count == 0 || 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` `                ``// less than the current element` `                ``while` `(s.Count > 0 && arr[i] < arr[s.Peek()]){` `                    ``int` `index = s.Pop();` `                    ``// update the minimum element` `                    ``if` `(arr[index] < arr[min]) {` `                        ``min = index;` `                    ``}` `                ``}` `                ``// push the current index onto the stack` `                ``s.Push(i);` `            ``}` `        ``}` `    `  `        ``// the minimum element is the element corresponding to` `        ``// the last index remaining on the stack` `        ``while` `(s.Count > 0){` `            ``int` `index = s.Pop();` `            ``if` `(arr[index] < arr[min]) {` `                ``min = index;` `            ``}` `        ``}` `    `  `        ``return` `arr[min];` `    ``}` `    `  `    ``static` `void` `Main(``string``[] args)` `    ``{` `    `  `        ``int``[] arr = { 50, 20, 3, 1, 6, 7, 10, 56 };` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(``"The minimum element is "` `+ findMin(arr, n));` `    `  `    ``}` `}`

## Javascript

 `function` `findMin(arr, n) {` `    ``const s = [];` `    ``let min_index = 0;`   `    ``// Traverse the array from left to right` `    ``for` `(let i = 0; i < n; i++) {` `        ``// Push the index onto the stack if the element is` `        ``// greater than or equal to the previous element` `        ``if` `(!s.length || arr[i] >= arr[s[s.length - 1]]) {` `            ``s.push(i);` `        ``} ``else` `{` `            ``// Pop all the indices from the stack until we` `            ``// find an index whose corresponding element is` `            ``// less than the current element` `            ``while` `(s.length && arr[i] < arr[s[s.length - 1]]) {` `                ``const index = s.pop();` `                ``// Update the minimum element` `                ``if` `(arr[index] < arr[min_index]) {` `                    ``min_index = index;` `                ``}` `            ``}` `            ``// Push the current index onto the stack` `            ``s.push(i);` `        ``}` `    ``}`   `    ``// The minimum element is the element corresponding to` `    ``// the last index remaining on the stack` `    ``while` `(s.length) {` `        ``const index = s.pop();` `        ``if` `(arr[index] < arr[min_index]) {` `            ``min_index = index;` `        ``}` `    ``}`   `    ``return` `arr[min_index];` `}`   `// Driver code` `const arr = [50, 20, 3, 1, 6, 7, 10, 56];` `const n = arr.length;` `console.log(``"The minimum element is"``, findMin(arr, n));`   `// This code is Contributed By - Dwaipayan Bandyopadhyay`

Output

```The maximum element is 1

```

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