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Find Minimum in Rotated Sorted Array II in C++
Suppose we have an array that is sorted, now that is rotated at some pivot. The pivot is not known before. We have to find the minimum element from that array. So if the array is like [4,5,5,5,6,8,2,3,4], then the minimum element is 2.
To solve this, we will follow these steps −
Define one method called search(), this takes arr, low and high
if low = high, then return arr[low]
mid := low + (high – low) / 2
ans := inf
if arr[low] < arr[mid], then ans := min of arr[low] and search(arr, mid, high)
otherwise when arr[high] > arr[mid], then ans := min of arr[mid] and search(arr, low, mid)
otherwise when arr[low] = arr[mid], then ans := min of arr[low] and search(arr, low + 1, high)
otherwise when arr[high] = arr[mid], then ans := min of arr[high] and search(arr, low, high - 1)
return ans
From the main method call solve(nums, 0, size of nums – 1)
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int search(vector <int>& arr, int low, int high){
if(low == high){
return arr[low];
}
int mid = low + (high - low) / 2;
int ans = INT_MAX;
if(arr[low] < arr[mid]){
ans = min(arr[low], search(arr, mid, high));
}
else if (arr[high] > arr[mid]){
ans = min(arr[mid], search(arr, low, mid));
}
else if(arr[low] == arr[mid]){
ans = min(arr[low], search(arr, low + 1, high));
}
else if(arr[high] == arr[mid]){
ans = min(arr[high], search(arr, low, high - 1));
}
return ans;
}
int findMin(vector<int>& nums) {
return search(nums, 0, nums.size() - 1);
}
};
main(){
Solution ob;
vector<int> v = {4,5,5,5,6,8,2,3,4};
cout <<(ob.findMin(v));
}Input
[4,5,5,5,6,8,2,3,4]
Output
2
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