Given a binary array arr[] of size N, and an integer K, the task is to calculate the number of ways to partition the array into non-overlapping subarrays, where each subarray has exactly K number 0s.
Examples:
Input: arr[] = [ 0, 0, 1, 1, 0, 1, 0], K = 2
Output: 3
Explanation: Different possible partitions are:
{{0, 0}, {1, 1, 0, 1, 0}}, {{0, 0, 1}, {1, 0, 1, 0}}, {{0, 0, 1, 1}, {0, 1, 0}}. So, the output will be 3.Input: arr[] = {0, 0, 1, 0, 1, 0}, K = 2
Output: 2Input: arr[] = [1, 0, 1, 1], K = 2
Output: 0
Approach: The approach to solve the problem is based on the following idea:
If jth 0 is the last 0 for a subarray and (j+1)th 0 is the first 0 of another subarray, then the possible number of ways to partition into those two subarrays is one more than the number of 1s in between jth and (j+1)th 0.
From the above observation, it can be said that the total possible ways to partition the subarray is the multiplication of the count of 1s between K*x th and (K*x + 1)th 0, for all possible x such that K*x does not exceed the total count of 0s in the array.
Follow the illustration below for a better understanding of the problem,
Illustration:
Consider array arr[] = {0, 0, 1, 1, 0, 1, 0, 1, 0, 0}, K = 2
Index of 2nd 0 and 3rd 0 are 1 and 4
=> Total number of 1s in between = 2.
=> Possible partition with these 0s = 2 + 1 = 3.
=> Total possible partitions till now = 3Index of 4th 0 and 5th 0 are 6 and 8
=> Total number of 1s in between = 1.
=> Possible partition with these 0s = 1 + 1 = 2.
=> Total possible partitions till now = 3*2 = 6The possible partitions are 6:
{{0, 0}, {1, 1, 0, 1, 0}, {1, 0, 0}}, {{0, 0}, {1, 1, 0, 1, 0, 1}, {0, 0}},
{{0, 0, 1}, {1, 0, 1, 0}, {1, 0, 0}}, {{0, 0, 1}, {1, 0, 1, 0, 1}, {0, 0}},
{{0, 0, 1, 1}, {0, 1, 0}, {1, 0, 0}}, {{0, 0, 1, 1}, {0, 1, 0, 1}, {0, 0}}
Follow the steps mentioned below to solve the problem:
- Initialize a counter variable to 1(claiming there exists at least one such possible way).
- If there are less than K 0s or number of 0s is not divisible by K, then such partition is not possible.
- Then, for every possible (K*x)th and (K*x + 1)th number of 0, calculate the number of possible partitions using the above observation and multiply that with the counter variable to get the total possible partitions.
- Return the value of the counter variable.
Here is the code for the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function used to calculate the number of// ways to divide the arrayint number_of_ways(vector<int>& arr, int K){ // Initialize a counter variable no_0 to // calculate the number of 0 int no_0 = 0; // Initialize a vector to // store the indices of 0s vector<int> zeros; for (int i = 0; i < arr.size(); i++) { if (arr[i] == 0) { no_0++; zeros.push_back(i); } } // If number of 0 is not divisible by K // or no 0 in the sequence return 0 if (no_0 % K || no_0 == 0) return 0; int res = 1; // For every (K*n)th and (K*n+1)th 0 // calculate the distance between them for (int i = K; i < zeros.size();) { res *= (zeros[i] - zeros[i - 1]); i += K; } // Return the number of such partitions return res;}// Driver codeint main(){ vector<int> arr = { 0, 0, 1, 1, 0, 1, 0 }; int K = 2; // Function call cout << number_of_ways(arr, K) << endl; return 0;} |
Java
// Java program for above approachimport java.io.*;import java.util.*;class GFG { // Function used to calculate the number of // ways to divide the array public static int number_of_ways(int arr[], int K) { // Initialize a counter variable no_0 to // calculate the number of 0 int no_0 = 0; // Initialize a arraylist to // store the indices of 0s ArrayList<Integer> zeros = new ArrayList<Integer>(); for (int i = 0; i < arr.length; i++) { if (arr[i] == 0) { no_0++; zeros.add(i); } } // If number of 0 is not divisible by K // or no 0 in the sequence return 0 if ((no_0 % K != 0) || no_0 == 0) return 0; int res = 1; // For every (K*n)th and (K*n+1)th 0 // calculate the distance between them for (int i = K; i < zeros.size();) { res *= (zeros.get(i) - zeros.get(i - 1)); i += K; } // Return the number of such partitions return res; } // Driver Code public static void main(String[] args) { int arr[] = { 0, 0, 1, 1, 0, 1, 0 }; int K = 2; // Function call System.out.println(number_of_ways(arr, K)); }}// This code is contributed by Rohit Pradhan |
Python3
# Python3 program for above approach# Function used to calculate the number of# ways to divide the arraydef number_of_ways(arr, K): # Initialize a counter variable no_0 to # calculate the number of 0 no_0 = 0 # Initialize am array to # store the indices of 0s zeros = [] for i in range(len(arr)): if arr[i] == 0: no_0 += 1 zeros.append(i) # If number of 0 is not divisible by K # or no 0 in the sequence return 0 if no_0 % K or no_0 == 0: return 0 res = 1 # For every (K*n)th and (K*n+1)th 0 # calculate the distance between them i = K while (i < len(zeros)): res *= (zeros[i] - zeros[i - 1]) i += K # Return the number of such partitions return res# Driver codearr = [0, 0, 1, 1, 0, 1, 0]K = 2# Function callprint(number_of_ways(arr, K))# This code is contributed by phasing17. |
C#
// C# program for above approachusing System;using System.Collections.Generic;public class GFG { // Function used to calculate the number of // ways to divide the array public static int number_of_ways(int[] arr, int K) { // Initialize a counter variable no_0 to // calculate the number of 0 int no_0 = 0; // Initialize a arraylist to // store the indices of 0s var zeros = new List<int>(); for (int i = 0; i < arr.Length; i++) { if (arr[i] == 0) { no_0++; zeros.Add(i); } } // If number of 0 is not divisible by K // or no 0 in the sequence return 0 if ((no_0 % K != 0) || no_0 == 0) return 0; int res = 1; // For every (K*n)th and (K*n+1)th 0 // calculate the distance between them for (int i = K; i < zeros.Count;) { res *= (zeros[i] - zeros[i - 1]); i += K; } // Return the number of such partitions return res; } public static void Main(string[] args) { int[] arr = { 0, 0, 1, 1, 0, 1, 0 }; int K = 2; // Function call Console.WriteLine(number_of_ways(arr, K)); }}// this code was contributed by phasing17 |
Javascript
<script> // JavaScript program for above approach // Function used to calculate the number of // ways to divide the array const number_of_ways = (arr, K) => { // Initialize a counter variable no_0 to // calculate the number of 0 let no_0 = 0; // Initialize a vector to // store the indices of 0s let zeros = []; for (let i = 0; i < arr.length; i++) { if (arr[i] == 0) { no_0++; zeros.push(i); } } // If number of 0 is not divisible by K // or no 0 in the sequence return 0 if (no_0 % K || no_0 == 0) return 0; let res = 1; // For every (K*n)th and (K*n+1)th 0 // calculate the distance between them for (let i = K; i < zeros.length;) { res *= (zeros[i] - zeros[i - 1]); i += K; } // Return the number of such partitions return res; } // Driver code let arr = [0, 0, 1, 1, 0, 1, 0]; let K = 2; // Function call document.write(number_of_ways(arr, K));// This code is contributed by rakeshsahni</script> |
3
Time Complexity: O(N)
Auxiliary Space: O(N)
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