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 become partitioned, i.e., it has first 0s than 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 contain all 0’s and right to point should contain 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 ith 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.
Implementation:
C++
// 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; } |
Java
// Java program to find minimum // toggle required import java.io.*; class GFG { // Function to calculate minimum toggling // required by using Dynamic programming static int minToggle(int arr[], int n) { int zero[] = new int[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 = Math.min(ans, i - zero[i] + zero[n] - zero[i]); return ans; } // Driver Program public static void main(String[] args) { int arr[] = { 1, 0, 1, 1, 0 }; int n = arr.length; System.out.println(minToggle(arr, n)); } } // This code is contributed by vt_m. |
Python3
# Python program to find # minimum toggle required # Function to calculate # minimum toggling # required by using # Dynamic programming def minToggle(arr, n): zero =[0 for i in range(n + 1+1)] zero[0] = 0 # Fill entries in zero[] # such that zero[i] # stores count of zeroes # to the left of i # (exl for i in range(1, n + 1): # 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) ans = n for i in range(1, n + 1): ans = min(ans, i - zero[i] + zero[n] - zero[i]) return ans # Driver Program arr = [1, 0, 1, 1, 0] n = len(arr) print(minToggle(arr, n)) # This code is contributed # by Anant Agarwal. |
C#
// C# program to find minimum // toggle required using System; class GFG { // Function to calculate minimum toggling // required by using Dynamic programming static int minToggle(int[] arr, int n) { int[] zero = new int[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 = Math.Min(ans, i - zero[i] + zero[n] - zero[i]); return ans; } // Driver Program public static void Main() { int[] arr = { 1, 0, 1, 1, 0 }; int n = arr.Length; Console.WriteLine(minToggle(arr, n)); } } // This code is contributed by Sam007. |
PHP
<?php // php program to find minimum toggle required // Function to calculate minimum toggling // required by using Dynamic programming function minToggle($arr, $n) { $zero[0] = 0; $zero[$n + 1]=0; // Fill entries in zero[] such that zero[i] // stores count of zeroes to the left of i // (exl for ($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) $ans = $n; for ($i = 1; $i <= $n; ++$i) $ans = min($ans, $i - $zero[$i] + $zero[$n] - $zero[$i]); return $ans; } // Driver Program $arr = array( 1, 0, 1, 1, 0 ); $n = sizeof($arr); echo minToggle($arr, $n) , "\n"; // This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript program to find minimum // toggle required // Function to calculate minimum toggling // required by using Dynamic programming function minToggle(arr, n) { let zero = new Array(n + 1); zero[0] = 0; // Fill entries in zero[] such that // zero[i] stores count of zeroes // to the left of i (exl for (let 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) let ans = n; for (let i = 1; i <= n; ++i) ans = Math.min(ans, i - zero[i] + zero[n] - zero[i]); return ans; } let arr = [ 1, 0, 1, 1, 0 ]; let n = arr.length; document.write(minToggle(arr, n)); </script> |
Output
2
Time complexity: O(n)
Auxiliary space: O(n)
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