Given an array arr[0 .. n-1] of **distinct** integers, the task is to find a local minima in it. We say that an element arr[x] is a local minimum if it is less than or equal to both its neighbors.

- For corner elements, we need to consider only one neighbor for comparison.
- There can be more than one local minima in an array, we need to find any one of them.

Examples:

Input: arr[] = {9, 6, 3, 14, 5, 7, 4}; Output: Index of local minima is 2 The output prints index of 3 because it is smaller than both of its neighbors. Note that indexes of elements 5 and 4 are also valid outputs. Input: arr[] = {23, 8, 15, 2, 3}; Output: Index of local minima is 1 Input: arr[] = {1, 2, 3}; Output: Index of local minima is 0 Input: arr[] = {3, 2, 1}; Output: Index of local minima is 2

A **simple solution **is to do a linear scan of array and as soon as we find a local minima, we return it. The worst case time complexity of this method would be O(n).

An **efficient solution** is based on Binary Search. We compare middle element with its neighbors. If middle element is not greater than any of its neighbors, then we return it. If the middle element is greater than its left neighbor, then there is always a local minima in left half (Why? take few examples). If the middle element is greater than its right neighbor, then there is always a local minima in right half (due to same reason as left half).

Below is the implementation of the above idea :

## C++

`// A C++ program to find a local minima in an array` `#include <stdio.h>` ` ` `// A binary search based function that returns` `// index of a local minima.` `int` `localMinUtil(` `int` `arr[], ` `int` `low, ` `int` `high, ` `int` `n)` `{` ` ` `// Find index of middle element` ` ` `int` `mid = low + (high - low)/2; ` `/* (low + high)/2 */` ` ` ` ` `// Compare middle element with its neighbours` ` ` `// (if neighbours exist)` ` ` `if` `((mid == 0 || arr[mid-1] > arr[mid]) &&` ` ` `(mid == n-1 || arr[mid+1] > arr[mid]))` ` ` `return` `mid;` ` ` ` ` `// If middle element is not minima and its left` ` ` `// neighbour is smaller than it, then left half` ` ` `// must have a local minima.` ` ` `else` `if` `(mid > 0 && arr[mid-1] < arr[mid])` ` ` `return` `localMinUtil(arr, low, (mid -1), n);` ` ` ` ` `// If middle element is not minima and its right` ` ` `// neighbour is smaller than it, then right half` ` ` `// must have a local minima.` ` ` `return` `localMinUtil(arr, (mid + 1), high, n);` `}` ` ` `// A wrapper over recursive function localMinUtil()` `int` `localMin(` `int` `arr[], ` `int` `n)` `{` ` ` `return` `localMinUtil(arr, 0, n-1, n);` `}` ` ` `/* Driver program to check above functions */` `int` `main()` `{` ` ` `int` `arr[] = {4, 3, 1, 14, 16, 40};` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `printf` `(` `"Index of a local minima is %d"` `,` ` ` `localMin(arr, n));` ` ` `return` `0;` `}` |

## Java

`// A Java program to find a local minima in an array` `import` `java.io.*;` ` ` `class` `GFG ` `{` ` ` ` ` `// A binary search based function that returns` ` ` `// index of a local minima.` ` ` `public` `static` `int` `localMinUtil(` `int` `[] arr, ` `int` `low, ` ` ` `int` `high, ` `int` `n)` ` ` `{` ` ` ` ` `// Find index of middle element` ` ` `int` `mid = low + (high - low) / ` `2` `;` ` ` ` ` `// Compare middle element with its neighbours` ` ` `// (if neighbours exist)` ` ` `if` `(mid == ` `0` `|| arr[mid - ` `1` `] > arr[mid] && mid == n - ` `1` `|| ` ` ` `arr[mid] < arr[mid + ` `1` `])` ` ` `return` `mid;` ` ` ` ` `// If middle element is not minima and its left` ` ` `// neighbour is smaller than it, then left half` ` ` `// must have a local minima.` ` ` `else` `if` `(mid > ` `0` `&& arr[mid - ` `1` `] < arr[mid])` ` ` `return` `localMinUtil(arr, low, mid - ` `1` `, n);` ` ` ` ` `// If middle element is not minima and its right` ` ` `// neighbour is smaller than it, then right half` ` ` `// must have a local minima.` ` ` `return` `localMinUtil(arr, mid + ` `1` `, high, n);` ` ` `}` ` ` ` ` `// A wrapper over recursive function localMinUtil()` ` ` `public` `static` `int` `localMin(` `int` `[] arr, ` `int` `n)` ` ` `{` ` ` `return` `localMinUtil(arr, ` `0` `, n - ` `1` `, n);` ` ` `}` ` ` ` ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{` ` ` ` ` `int` `arr[] = {` `4` `, ` `3` `, ` `1` `, ` `14` `, ` `16` `, ` `40` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(` `"Index of a local minima is "` `+ localMin(arr, n));` ` ` ` ` `}` `}` ` ` `//This code is contributed by Dheerendra Singh` |

## Python3

`# Python3 program to find a` `# local minima in an array` ` ` `# A binary search based function that ` `# returns index of a local minima.` `def` `localMinUtil(arr, low, high, n):` ` ` ` ` `# Find index of middle element` ` ` `mid ` `=` `low ` `+` `(high ` `-` `low) ` `/` `/` `2` ` ` ` ` `# Compare middle element with its ` ` ` `# neighbours (if neighbours exist)` ` ` `if` `(mid ` `=` `=` `0` `or` `arr[mid ` `-` `1` `] > arr[mid] ` `and` ` ` `mid ` `=` `=` `n ` `-` `1` `or` `arr[mid] < arr[mid ` `+` `1` `]):` ` ` `return` `mid` ` ` ` ` `# If middle element is not minima and its left` ` ` `# neighbour is smaller than it, then left half` ` ` `# must have a local minima.` ` ` `elif` `(mid > ` `0` `and` `arr[mid ` `-` `1` `] < arr[mid]):` ` ` `return` `localMinUtil(arr, low, mid ` `-` `1` `, n)` ` ` ` ` `# If middle element is not minima and its right` ` ` `# neighbour is smaller than it, then right half` ` ` `# must have a local minima.` ` ` `return` `localMinUtil(arr, mid ` `+` `1` `, high, n)` ` ` `# A wrapper over recursive function localMinUtil()` `def` `localMin(arr, n):` ` ` ` ` `return` `localMinUtil(arr, ` `0` `, n ` `-` `1` `, n)` ` ` `# Driver code` `arr ` `=` `[` `4` `, ` `3` `, ` `1` `, ` `14` `, ` `16` `, ` `40` `]` `n ` `=` `len` `(arr)` `print` `(` `"Index of a local minima is "` `,` ` ` `localMin(arr, n))` ` ` `# This code is contributed by Anant Agarwal.` |

## C#

`// A C# program to find a ` `// local minima in an array` `using` `System;` ` ` `class` `GFG ` `{` ` ` ` ` `// A binary search based function that returns` ` ` `// index of a local minima.` ` ` `public` `static` `int` `localMinUtil(` `int` `[] arr, ` `int` `low, ` ` ` `int` `high, ` `int` `n)` ` ` `{` ` ` ` ` `// Find index of middle element` ` ` `int` `mid = low + (high - low) / 2;` ` ` ` ` `// Compare middle element with its neighbours` ` ` `// (if neighbours exist)` ` ` `if` `(mid == 0 || arr[mid - 1] > arr[mid] && ` ` ` `mid == n - 1 || arr[mid] < arr[mid + 1])` ` ` `return` `mid;` ` ` ` ` `// If middle element is not minima and its left` ` ` `// neighbour is smaller than it, then left half` ` ` `// must have a local minima.` ` ` `else` `if` `(mid > 0 && arr[mid - 1] < arr[mid])` ` ` `return` `localMinUtil(arr, low, mid - 1, n);` ` ` ` ` `// If middle element is not minima and its right` ` ` `// neighbour is smaller than it, then right half` ` ` `// must have a local minima.` ` ` `return` `localMinUtil(arr, mid + 1, high, n);` ` ` `}` ` ` ` ` `// A wrapper over recursive function localMinUtil()` ` ` `public` `static` `int` `localMin(` `int` `[] arr, ` `int` `n)` ` ` `{` ` ` `return` `localMinUtil(arr, 0, n - 1, n);` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main () ` ` ` `{` ` ` ` ` `int` `[]arr = {4, 3, 1, 14, 16, 40};` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(` `"Index of a local minima is "` `+` ` ` `localMin(arr, n));` ` ` ` ` `}` `}` ` ` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// A PHP program to find a ` `// local minima in an array` ` ` `// A binary search based ` `// function that returns` `// index of a local minima.` `function` `localMinUtil(` `$arr` `, ` `$low` `, ` `$high` `, ` `$n` `)` `{` ` ` ` ` `// Find index of middle element` ` ` `/* (low + high)/2 */` ` ` `$mid` `= ` `$low` `+ (` `$high` `- ` `$low` `) / 2; ` ` ` ` ` `// Compare middle element` ` ` `// with its neighbours` ` ` `// (if neighbours exist)` ` ` `if` `((` `$mid` `== 0 ` `or` `$arr` `[` `$mid` `- 1] > ` `$arr` `[` `$mid` `]) ` `and` ` ` `(` `$mid` `== ` `$n` `- 1 ` `or` `$arr` `[` `$mid` `+ 1] > ` `$arr` `[` `$mid` `]))` ` ` `return` `$mid` `;` ` ` ` ` `// If middle element is not` ` ` `// minima and its left` ` ` `// neighbour is smaller than` ` ` `// it, then left half` ` ` `// must have a local minima.` ` ` `else` `if` `(` `$mid` `> 0 ` `and` `$arr` `[` `$mid` `- 1] < ` `$arr` `[` `$mid` `])` ` ` `return` `localMinUtil(` `$arr` `, ` `$low` `, (` `$mid` `- 1), ` `$n` `);` ` ` ` ` `// If middle element is not` ` ` `// minima and its right` ` ` `// neighbour is smaller than` ` ` `// it, then right half` ` ` `// must have a local minima.` ` ` `return` `localMinUtil(arr, (mid + 1), high, n);` `}` ` ` `// A wrapper over recursive ` `// function localMinUtil()` `function` `localMin( ` `$arr` `, ` `$n` `)` `{` ` ` `return` `floor` `(localMinUtil(` `$arr` `, 0, ` `$n` `- 1, ` `$n` `));` `}` ` ` ` ` `// Driver Code` ` ` `$arr` `= ` `array` `(4, 3, 1, 14, 16, 40);` ` ` `$n` `= ` `count` `(` `$arr` `);` ` ` `echo` `"Index of a local minima is "` `,` ` ` `localMin(` `$arr` `, ` `$n` `);` ` ` `// This code is contributed by anuj_67.` `?>` |

Output:

Index of a local minima is 2

Time Complexity : O(Log n)

Related Problem :**Find a peak element**

This article is contributed by **Roshni Agarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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