Given a number N. The task is to find the minimum number of binary strings required to represent the given number as the sum of the binary strings.
Examples:
Input : 131
Output : Minimum Number of binary strings needed: 3
111 10 10
Input : 564
Output :Minimum Number of binary strings needed: 6
111 111 111 111 110 10
Approach:
- Store all digits of the given number in the array.
- Find the maximum digit in the array. This maximum number(maxi) indicates the number of binary strings required to represent the given number.
- Now, find maxi numbers by substituting 0’s and 1’s greedily.
Below is the implementation of the above approach:
C++
// C++ program to find the minimum number of // binary strings to represent a number#include <bits/stdc++.h>using namespace std;// Function to find the minimum number of // binary strings to represent a numbervoid minBinary(int n){ int digit[10], len = 0; while (n > 0) { digit[len++] = n % 10; n /= 10; } // Storing digits in correct order reverse(digit, digit + len); int ans = 0; // Find the maximum digit in the array for (int i = 0; i < len; i++) { ans = max(ans, digit[i]); } cout << "Minimum Number of binary strings needed: " << ans << endl; // Traverse for all the binary strings for (int i = 1; i <= ans; i++) { int num = 0; for (int j = 0; j < len; j++) { // If digit at jth position is greater // than 0 then substitute 1 if (digit[j] > 0) { num = num * 10 + 1; digit[j]--; } else { num *= 10; } } cout << num << " "; }}// Driver codeint main(){ int n = 564; minBinary(n); return 0;} |
Java
// Java program to find the minimum number of // binary Strings to represent a numberimport java.util.*;class GFG { // Function to find the minimum number of // binary Strings to represent a number static void minBinary(int n) { int[] digit = new int[10]; int len = 0; while (n > 0) { digit[len++] = n % 10; n /= 10; } // Storing digits in correct order digit = reverse(digit, 0, len - 1); int ans = 0; // Find the maximum digit in the array for (int i = 0; i < len; i++) { ans = Math.max(ans, digit[i]); } System.out.print("Minimum Number of binary" + " Strings needed: " + ans + "\n"); // Traverse for all the binary Strings for (int i = 1; i <= ans; i++) { int num = 0; for (int j = 0; j < len; j++) { // If digit at jth position is greater // than 0 then substitute 1 if (digit[j] > 0) { num = num * 10 + 1; digit[j]--; } else { num *= 10; } } System.out.print(num + " "); } } static int[] reverse(int str[], int start, int end) { // Temporary variable to store character int temp; while (start <= end) { // Swapping the first and last character temp = str[start]; str[start] = str[end]; str[end] = temp; start++; end--; } return str; } // Driver code public static void main(String[] args) { int n = 564; minBinary(n); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to find the minimum number of # binary strings to represent a number# Function to find the minimum number of # binary strings to represent a numberdef minBinary(n): digit = [0 for i in range(3)] len = 0 while (n > 0): digit[len] = n % 10 len += 1 n //= 10 # Storing digits in correct order digit = digit[::-1] ans = 0 # Find the maximum digit in the array for i in range(len): ans = max(ans, digit[i]) print("Minimum Number of binary strings needed:", ans) # Traverse for all the binary strings for i in range(1, ans + 1, 1): num = 0 for j in range(0, len, 1): # If digit at jth position is greater # than 0 then substitute 1 if (digit[j] > 0): num = num * 10 + 1 digit[j] -= 1 else: num *= 10 print(num, end = " ")# Driver codeif __name__ == '__main__': n = 564 minBinary(n) # This code is contributed by# Surendra_Gangwar |
C#
// C# program to find the minimum number of // binary Strings to represent a numberusing System;class GFG { // Function to find the minimum number of // binary Strings to represent a number static void minBinary(int n) { int[] digit = new int[10]; int len = 0; while (n > 0) { digit[len++] = n % 10; n /= 10; } // Storing digits in correct order digit = reverse(digit, 0, len - 1); int ans = 0; // Find the maximum digit in the array for (int i = 0; i < len; i++) { ans = Math.Max(ans, digit[i]); } Console.Write("Minimum Number of binary" + " Strings needed: " + ans + "\n"); // Traverse for all the binary Strings for (int i = 1; i <= ans; i++) { int num = 0; for (int j = 0; j < len; j++) { // If digit at jth position is greater // than 0 then substitute 1 if (digit[j] > 0) { num = num * 10 + 1; digit[j]--; } else { num *= 10; } } Console.Write(num + " "); } } static int[] reverse(int []str, int start, int end) { // Temporary variable to store character int temp; while (start <= end) { // Swapping the first and // last character temp = str[start]; str[start] = str[end]; str[end] = temp; start++; end--; } return str; } // Driver code public static void Main(String[] args) { int n = 564; minBinary(n); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to // find the minimum number of // binary Strings to represent// a number // Function to find the minimum number of // binary Strings to represent a number function minBinary(n) { var digit = Array(10).fill(0); var len = 0; while (n > 0) { digit[len++] = n % 10; n = parseInt(n/10); } // Storing digits in correct order digit = reverse(digit, 0, len - 1); var ans = 0; // Find the maximum digit in the array for (i = 0; i < len; i++) { ans = Math.max(ans, digit[i]); } document.write("Minimum Number of binary" + " Strings needed: " + ans + "<br/>"); // Traverse for all the binary Strings for (i = 1; i <= ans; i++) { var num = 0; for (j = 0; j < len; j++) { // If digit at jth position is greater // than 0 then substitute 1 if (digit[j] > 0) { num = num * 10 + 1; digit[j]--; } else { num *= 10; } } document.write(num + " "); } } function reverse(str , start , end) { // Temporary variable to store character var temp; while (start <= end) { // Swapping the first and last character temp = str[start]; str[start] = str[end]; str[end] = temp; start++; end--; } return str; } // Driver code var n = 564; minBinary(n);// This code contributed by umadevi9616 </script> |
Output:
Minimum No of binary strings needed: 6 111 111 111 111 110 10
Time Complexity: O(N)
Auxiliary Space: O(1)
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