# Find local minima in an array

## Problem

Given an array of unique integers whose first two numbers are decreasing and last two numbers are increasing, find a number in the array which is local minima. A number in the array is called local minima if it is smaller than both its left and right numbers. For example in the array 9,7,2,8,5,6,3,4

2 is a local minima as it is smaller than its left and right number 7 and 8. Similarly 5 is another local minima as it is between 8 and 6, both larger than 5. You need to find any one of the local minima.

## Solution

First we need to understand that if in an array of unique integers first two numbers are decreasing and last two numbers are increasing there ought to be a local minima. Why so? We can prove it in two ways. First we will do it by negation. If first two numbers are decreasing, and there is no local minima, that means 3rd number is less than 2nd number. otherwise 2nd number would have been local minima. Following the same logic 4th number will have to be less than 3rd number and so on and so forth. So the numbers in the array will have to be in decreasing order. Which violates the constraint of last two numbers being in increasing order. This proves by negation that there need to be a local minima.

We can prove this in some other way also. Suppose we represent the array as a 2-D graph where the index of the numbers in the array represents the x-coordinate. and the number represents the y-coordinate. Now for the first two numbers, derivative will be negative, and for last two numbers derivative will be positive. So at some point the derivative line will have to cross the x axis. As the array contains only unique elements there cannot be a derivative point on the x axis. Because that will mean that two consecutive index having same number. So for any intersection of x axis by the derivative line will be a local minima.

We will solve this problem in O(log n) time by divide and conquer method. We will first check the mid index of the array. If it is smaller than its left and right, then it is the answer. If it is bigger than the left number then from start to left we have a subproblem, and as we proved already that starting with decreasing and ending with increasing sequence array will have to have a local minima, we can safely go to the left subarray. Otherwise if mid is bigger than its right, then we go to the right subarray. This way we guarantee a O(log n) algorithm to find any of the local minima present in the array.

## Code

public class LocalMinima { public static void main(String[] args) { int[] arr = {64, 14, 52, 27, 71, 19, 63, 1, 16, 57}; for (int num : arr) System.out.print(num + ", "); System.out.println(); int minima = findLocalMinima(arr); System.out.println(minima); } private static int findLocalMinima(int[] arr) { return findLocalMinima(arr, 0, arr.length); } private static int findLocalMinima(int[] arr, int start, int end) { int mid = (start + end) / 2; if (mid - 2 < 0 && mid + 1 >= arr.length) return -1; if (arr[mid - 2] > arr[mid - 1] && arr[mid - 1] < arr[mid]) return arr[mid - 1]; if (arr[mid - 1] > arr[mid - 2]) return findLocalMinima(arr, start, mid); else return findLocalMinima(arr, mid, end); } }