Given two integers M and K, the task is to find the Kth smallest number with digit sum as M.
Examples:
Input: M = 5, K = 3
Output: 23
Sequence of numbers starting from 1 with digit sum as 5 is as follows:
5
14
23
32
41
So 3rd smallest number is 23
Input: M = 4, K = 1
Output: 4
Approach: We need to find the sequence of numbers starting from 1 with the sum of digits as m.
- Write a recursive function that will increase the numbers until the sum of the digits of the number will be equal to our required sum M.
- For that, start from 0 always and check for single digit number that will add upto M
- Let’s take an example for sum = 5 so we will start from 0 and go upto 5 in single digit as 6 exceeds out required sum
- Now from 5 we will move to two digits number 10 and we will go up to 14 because the sum of digits of 14 is 5 and 15 will exceed the required sum and so on, then further we will move in 20’s and this goes on up to 50 because after 50 till 99 the sum of digits of every number will be greater than 5 so we will skip that.
- Now we will move in three digits 100, 101, 102, … and similarly this operation will be going on until the sum of digits of the number are equal to 15.
- We will keep insert the elements into set whose sum of digits is equals to M.
for(int i=0;i<=min(left, 9LL);i++){
dfs(num*10+i, left-i, ct+1);
}
To find kth smallest we need to sort the sequence, so it will be better if we store the numbers in a set in C++, as numbers in a set are arranged in sorted order.
Note that this approach will not work for bigger input values.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 2005;set<int> ans;// Recursively moving to add// all the numbers upto a limit// with sum of digits as mvoid dfs(int num, int left, int ct){ // Max number of digits allowed in // a number for this implementation if (ct >= 15) return; if (left == 0) ans.insert(num); for (int i = 0; i <= min(left, 9LL); i++) dfs(num * 10 + i, left - i, ct + 1);}// Function to return the kth number// with sum of digits as mint getKthNum(int m, int k){ dfs(0, m, 0); int ct = 0; for (auto it : ans) { ct++; // The kth smallest number is found if (ct == k) { return it; } } return -1;}// Driver codeint main(){ int m = 5, k = 3; cout << getKthNum(m, k); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static int N = 2005;static Set<Integer> ans = new LinkedHashSet<Integer>();// Recursively moving to add// all the numbers upto a limit// with sum of digits as mstatic void dfs(int num, int left, int ct){ // Max number of digits allowed in // a number for this implementation if (ct >= 15) return; if (left == 0) ans.add(num); for (int i = 0; i <= Math.min(left, 9); i++) dfs(num * 10 + i, left - i, ct + 1);}// Function to return the kth number// with sum of digits as mstatic int getKthNum(int m, int k){ dfs(0, m, 0); int ct = 0; for (int it : ans) { ct++; // The kth smallest number is found if (ct == k) { return it; } } return -1;}// Driver codepublic static void main(String[] args){ int m = 5, k = 3; System.out.println(getKthNum(m, k));}} // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach# define long longN = 2005ans = dict()# Recursively moving to add# all the numbers upto a limit# with sum of digits as mdef dfs(n, left, ct): # Max number of digits allowed in # a number for this implementation if (ct >= 15): return if (left == 0): ans[n] = 1 for i in range(min(left, 9) + 1): dfs(n * 10 + i, left - i, ct + 1)# Function to return the kth number# with sum of digits as mdef getKthNum(m, k): dfs(0, m, 0) c = 0 for it in sorted(ans.keys()): c += 1 # The kth smallest number is found if (c == k): return it return -1# Driver codem = 5k = 3print(getKthNum(m, k))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {static int N = 2005;static HashSet<int> ans = new HashSet<int>();// Recursively moving to add// all the numbers upto a limit// with sum of digits as mstatic void dfs(int num, int left, int ct){ // Max number of digits allowed in // a number for this implementation if (ct >= 15) return; if (left == 0) ans.Add(num); for (int i = 0; i <= Math.Min(left, 9); i++) dfs(num * 10 + i, left - i, ct + 1);}// Function to return the kth number// with sum of digits as mstatic int getKthNum(int m, int k){ dfs(0, m, 0); int ct = 0; foreach (int it in ans) { ct++; // The kth smallest number is found if (ct == k) { return it; } } return -1;}// Driver codepublic static void Main(String[] args){ int m = 5, k = 3; Console.WriteLine(getKthNum(m, k));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approachlet N = 2005;let ans = new Set();// Recursively moving to add// all the numbers upto a limit// with sum of digits as mfunction dfs(num,left,ct){ // Max number of digits allowed in // a number for this implementation if (ct >= 15) return; if (left == 0) ans.add(num); for (let i = 0; i <= Math.min(left, 9); i++) dfs(num * 10 + i, left - i, ct + 1);}// Function to return the kth number// with sum of digits as mfunction getKthNum(m,k){ dfs(0, m, 0); let ct = 0; for (let it of ans.values()) { ct++; // The kth smallest number is found if (ct == k) { return it; } } return -1;}// Driver codelet m = 5, k = 3; document.write(getKthNum(m, k));// This code is contributed by unknown2108</script> |
Output:
23
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