Given an array of n integers containing only 0 and 1. Find the minimum toggles (switch from 0 to 1 or vice-versa) required such the array the array become partitioned, i.e., it has first 0s then 1s. There should be at least one 0 in the beginning, and there can be zero or more 1s in the end.

Input: arr[] = {1, 0, 1, 1, 0} Output: 2 Toggle the first and last element i.e., 1 -> 0 0 -> 1 Final array will become: arr[] = {0 0 1 1 1} Since first two consecutive elements are all 0s and rest three consecutive elements are all 1s. Therefore minimum two toggles are required. Input: arr[] = {0, 1, 0, 0, 1, 1, 1} Output: 1 Input: arr[] = {1, 1} Output: 1 There should be at least one 0. Input: arr[] = {0, 0} Output: 0 There can zero 1s.

If we observe the question then we will find that there will definitely exist a point from 0 to n-1 where all elements left to that point should contains all 0’s and right to point should contains all 1’s. Those indices which don’t obey this law will have to be removed. The idea is to count all 0s from left to right.

Let zero[i] denotes the number of 0's till i^{th}index, then for each i, minimum number of toggles required can be written as:i - zero[i] + zero[n] - zero[i]. The part i - zero[i] indicates number of 1's to be toggled and the part zero[n] - zero[i] indicates number of 0's to be toggled. After that we just need to take minimum of all to get the final answer.

// C++ program to find minimum toggle required #include <bits/stdc++.h> using namespace std; // Function to calculate minimum toggling // required by using Dynamic programming int minToggle(int arr[], int n) { int zero[n+1]; zero[0] = 0; // Fill entries in zero[] such that zero[i] // stores count of zeroes to the left of i // (exl for (int i=1; i<=n; ++i) { // If zero found update zero[] array if (arr[i-1] == 0) zero[i] = zero[i-1] + 1; else zero[i] = zero[i-1]; } // Finding the minimum toggle required from // every index(0 to n-1) int ans = n; for (int i=1; i <= n; ++i) ans = min(ans, i-zero[i]+zero[n]-zero[i]); return ans; } // Driver Program int main() { int arr[] = {1, 0, 1, 1, 0}; int n = sizeof (arr) / sizeof (arr[0]); cout << minToggle(arr, n) << "\n"; return 0; }

**Output:**

2

**Time complexity: **O(n)

**Auxiliary space: **O(n)

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