Minimum numbers with one’s place as 9 to be added to get N
Given an integer N, the task is to find the minimum count of numbers which needs to be added to get N. (All such numbers must have 9 as one’s digit)
Examples:
Input: N = 27
Output: 3
27 = 9 + 9 + 9Input: N = 109
Output: 1
109 itself has 9 as one’s digit.
Approach:
- Check the one’s digit of N, based on one’s digit the minimum count of numbers which needs to be added can be easily found.
- If one’s digit is 9: The answer will 1 as the number itself has 9 as its one’s place digit.
- If one’s digit is:
- 1: 9 has to be added 9 times i.e. (9 * 9 = 81).
- 2: 9 has to be added 8 times i.e. (9 * 8 = 72).
- 3: 9 has to be added 7 times i.e. (9 * 7 = 63).
- 4: 9 has to be added 6 times i.e. (9 * 6 = 54).
- 5: 9 has to be added 5 times i.e. (9 * 5 = 45).
- 6: 9 has to be added 4 times i.e. (9 * 4 = 36).
- 7: 9 has to be added 3 times i.e. (9 * 3 = 27).
- 8: 9 has to be added 2 times i.e. (9 * 2 = 18).
- 0: 9 has to be added 10 times i.e. (9 * 10 = 90).
- Observation here is that only the minimum multiple count for all the cases above mentioned have to be added. This is because say for one’s digit as 4, all 9 (one’s place) from 6 numbers can be used and the result can be subtracted from N say it is M. Now, M will have 0 in one’s place. So, just use 5 numbers as 9 and sixth number as (M + 9).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find minimum count// of numbers(with one's digit 9)// that sum up to Nint findMin(int N){ // Fetch one's digit int digit = N % 10; // Apply Cases mentioned in approach switch (digit) { case 0: if (N >= 90) return 10; break; case 1: if (N >= 81) return 9; break; case 2: if (N >= 72) return 8; break; case 3: if (N >= 63) return 7; break; case 4: if (N >= 54) return 6; break; case 5: if (N >= 45) return 5; break; case 6: if (N >= 36) return 4; break; case 7: if (N >= 27) return 3; break; case 8: if (N >= 18) return 2; break; case 9: if (N >= 9) return 1; break; } // If no possible answer exists return -1;}// Driver codeint main(){ int N = 27; cout << findMin(N);} |
Java
// Java implementation of the approach class GFG{ // Function to find minimum count // of numbers(with one's digit 9) // that sum up to N static int findMin(int N) { // Fetch one's digit int digit = N % 10; // Apply Cases mentioned in approach switch (digit) { case 0: if (N >= 90) return 10; break; case 1: if (N >= 81) return 9; break; case 2: if (N >= 72) return 8; break; case 3: if (N >= 63) return 7; break; case 4: if (N >= 54) return 6; break; case 5: if (N >= 45) return 5; break; case 6: if (N >= 36) return 4; break; case 7: if (N >= 27) return 3; break; case 8: if (N >= 18) return 2; break; case 9: if (N >= 9) return 1; break; } // If no possible answer exists return -1; } // Driver code public static void main (String[] args) { int N = 27; System.out.println(findMin(N)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to find minimum count# of numbers(with one's digit 9)# that sum up to Ndef findMin(N: int): # Fetch one's digit digit = N % 10 # Apply Cases mentioned in approach if digit == 0 and N >= 90: return 10 elif digit == 1 and N >= 81: return 9 elif digit == 2 and N >= 72: return 8 elif digit == 3 and N >= 63: return 7 elif digit == 4 and N >= 54: return 6 elif digit == 5 and N >= 45: return 5 elif digit == 6 and N >= 36: return 4 elif digit == 7 and N >= 27: return 3 elif digit == 8 and N >= 18: return 2 elif digit == 9 and N >= 9: return 1 # If no possible answer exists return -1# Driver Codeif __name__ == "__main__": N = 27 print(findMin(N))# This code is contributed by# sanjeev2552 |
C#
// C# implementation of the approach using System; class GFG{ // Function to find minimum count // of numbers(with one's digit 9) // that sum up to N static int findMin(int N) { // Fetch one's digit int digit = N % 10; // Apply Cases mentioned in approach switch (digit) { case 0: if (N >= 90) return 10; break; case 1: if (N >= 81) return 9; break; case 2: if (N >= 72) return 8; break; case 3: if (N >= 63) return 7; break; case 4: if (N >= 54) return 6; break; case 5: if (N >= 45) return 5; break; case 6: if (N >= 36) return 4; break; case 7: if (N >= 27) return 3; break; case 8: if (N >= 18) return 2; break; case 9: if (N >= 9) return 1; break; } // If no possible answer exists return -1; } // Driver code public static void Main (String[] args) { int N = 27; Console.WriteLine(findMin(N)); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to find minimum count// of numbers(with one's digit 9)// that sum up to Nfunction findMin(N){ // Fetch one's digit let digit = N % 10; // Apply Cases mentioned in approach switch (digit) { case 0: if (N >= 90) return 10; break; case 1: if (N >= 81) return 9; break; case 2: if (N >= 72) return 8; break; case 3: if (N >= 63) return 7; break; case 4: if (N >= 54) return 6; break; case 5: if (N >= 45) return 5; break; case 6: if (N >= 36) return 4; break; case 7: if (N >= 27) return 3; break; case 8: if (N >= 18) return 2; break; case 9: if (N >= 9) return 1; break; } // If no possible answer exists return -1;}// Driver codelet N = 27;document.write(findMin(N));// This code is contributed by souravmahato348</script> |
Output:
3
Time Complexity: O(1)
Auxiliary Space: O(1)
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