Given an Integer N, and a number K, the task is to find out the total numbers from 0 to N which have exactly K non zero digits and the sum of those digits should be odd and that sum should be distinct. The number N can be as large as 10^18.
Examples:
Input : N = 10, K = 1
Output : 5
The numbers which follow the conditions are ->
1, 3, 5, 7 and 9
The digit sum of 10 that is (1+0) = 1 is also odd, but 1 is already included in our count.
Input : N = 100, K = 2
Output : 8
Prerequisites: Digit-DP
Naive Approach:
A naive approach for linear traversing in O(N) of all elements from 0 to N and calculating sum of digits in log(n) where n is the number of digits in that number will fail for large inputs of N.
Efficient Approach:
- We can use Dynamic Programming and it’s very useful technique that is digit-dp to solve this problem.
- So instead of keeping a record of non-zero integers, we keep a record of zeroes we can keep at different indices, idx in variable K. The number of zeroes we can keep can be found initially by subtracting K with the number of digits in N.
- We keep all the digits of N into a vector say, digits.
- Now, we calculate range of elements we can keep at index idx by analysing K.
- Suppose at index idx, we are left with K = 1 (A Non-zero value), then our range to put elements is [0, j] where j is the upper bound decided by the tight value obtained from the current index of the digit from vector digits.
- If at idx, we are left with K = 0, then our range becomes [1, j] because we can’t put in 0 there.
- Suppose at index idx, we are left with K = 1 (A Non-zero value), then our range to put elements is [0, j] where j is the upper bound decided by the tight value obtained from the current index of the digit from vector digits.
- Now, also take a parameter that is sum, which will calculate the sum of digits of a number till the base case hits successfully.
- Also, a boolean map is used which will store all the odd sums calculated already, so it gives distinct odd sums.
- The recurrence will be:
where j = digits[idx] if tight = 0, else j = 9
- Base Case:
- When idx = digits.size(), K == 0 and sum is odd.
We mark the sum as true and return 1 else return 0.
- If idx > digits.size() then return 0.
- When idx = digits.size(), K == 0 and sum is odd.
So we create a DP table say DP[idx][K][tight][sum] which will store our result from the recurrence above and return the count by memoizing it to this DP table.
Below is the implementation of the above approach:
C++
// C++ program to Count the numbers having// exactly K non-zero digits and sum// of digits are odd and distinct.#include <bits/stdc++.h>using namespace std;// To store digits of Nvector<int> digits;// visited mapbool vis[170] = { false };// DP Tableint dp[19][19][2][170];// Push all the digits of N into// digits vectorvoid ConvertIntoDigit(int n){ while (n) { int dig = n % 10; digits.push_back(dig); n /= 10; } reverse(digits.begin(), digits.end());}// Function returns the countint solve(int idx, int k, int tight, int sum){ // If desired number is formed // whose sum is odd if (idx == digits.size() && k == 0 && sum & 1) { // If it is not present in map, // mark it as true and return 1 if (!vis[sum]) { vis[sum] = 1; return 1; } // Sum is present in map already return 0; } // Desired result not found if (idx > digits.size()) { return 0; } // If that state is already calculated // just return that state value if (dp[idx][k][tight][sum]) { return dp[idx][k][tight][sum]; } // Upper limit int j; if (tight == 0) { j = digits[idx]; } else { j = 9; } // To store the count of // desired numbers int cnt = 0; // If k is non-zero, i ranges from // 0 to j else [1, j] for (int i = (k ? 0 : 1); i <= j; i++) { int newtight = tight; if (i < j) { newtight = 1; } // If current digit is 0, decrement // k and recurse sum is not changed // as we are just adding 0 that // makes no difference if (i == 0) cnt += solve(idx + 1, k - 1, newtight, sum); // If i is non zero, then k remains // unchanged and value is added to sum else cnt += solve(idx + 1, k, newtight, sum + i); } // Memoize and return return dp[idx][k][tight][sum] = cnt;}// Driver codeint main(){ // K is the number of exact non-zero // elements to have in number int N, k; N = 169, k = 2; // break N into its digits ConvertIntoDigit(N); // We keep record of 0s we need to // place in the number k = digits.size() - k; cout << solve(0, k, 0, 0);} |
Java
// Java program to count the numbers having// exactly K non-zero digits and sum// of digits are odd and distinct.import java.util.*;class GFG{// To store digits of Nstatic Vector<Integer> digits = new Vector<Integer>();// visited mapstatic boolean []vis = new boolean[170];// DP Tablestatic int [][][][]dp = new int[19][19][2][170];// Push all the digits of N into// digits vectorstatic void ConvertIntoDigit(int n){ while (n > 0) { int dig = n % 10; digits.add(dig); n /= 10; } Collections.reverse(digits);}// Function returns the countstatic int solve(int idx, int k, int tight, int sum){ // If desired number is formed // whose sum is odd if (idx == digits.size() && k == 0 && sum % 2 == 1) { // If it is not present in map, // mark it as true and return 1 if (!vis[sum]) { vis[sum] = true; return 1; } // Sum is present in map already return 0; } // Desired result not found if (idx > digits.size()) { return 0; } // If that state is already calculated // just return that state value if (dp[idx][k][tight][sum] > 0) { return dp[idx][k][tight][sum]; } // Upper limit int j; if (idx < digits.size() && tight == 0) { j = digits.get(idx); } else { j = 9; } // To store the count of // desired numbers int cnt = 0; // If k is non-zero, i ranges from // 0 to j else [1, j] for(int i = (k > 0 ? 0 : 1); i <= j; i++) { int newtight = tight; if (i < j) { newtight = 1; } // If current digit is 0, decrement // k and recurse sum is not changed // as we are just adding 0 that // makes no difference if (i == 0) cnt += solve(idx + 1, k - 1, newtight, sum); // If i is non zero, then k remains // unchanged and value is added to sum else cnt += solve(idx + 1, k, newtight, sum + i); } // Memoize and return return dp[idx][k][tight][sum] = cnt;}// Driver codepublic static void main(String[] args){ // K is the number of exact non-zero // elements to have in number int N, k; N = 169; k = 2; // break N into its digits ConvertIntoDigit(N); // We keep record of 0s we need to // place in the number k = digits.size() - k; System.out.print(solve(0, k, 0, 0));}}// This code is contributed by amal kumar choubey |
Python3
# Python3 program to Count the numbers having# exactly K non-zero digits and sum# of digits are odd and distinct. # To store digits of Ndigits = [] # visited mapvis = [False for i in range(170)] # DP Tabledp = [[[[0 for l in range(170)] for k in range(2)] for j in range(19)] for i in range(19)] # Push all the digits of N into# digits vectordef ConvertIntoDigit(n): while (n > 0): dig = n % 10; digits.append(dig); n //= 10; digits.reverse() # Function returns the countdef solve(idx, k, tight, sum): # If desired number is formed # whose sum is odd if (idx == len(digits) and k == 0 and sum % 2 == 1): # If it is not present in map, # mark it as true and return 1 if (not vis[sum]): vis[sum] = True; return 1; # Sum is present in map already return 0; # Desired result not found if (idx > len(digits)): return 0; # If that state is already calculated # just return that state value if (dp[idx][k][tight][sum]): return dp[idx][k][tight][sum]; # Upper limit j = 0; if (idx<len(digits) and tight == 0): j = digits[idx]; else: j = 9; # To store the count of # desired numbers cnt = 0; # If k is non-zero, i ranges from # 0 to j else [1, j] for i in range(0 if k else 1, j + 1): newtight = tight; if (i < j): newtight = 1; # If current digit is 0, decrement # k and recurse sum is not changed # as we are just adding 0 that # makes no difference if (i == 0): cnt += solve(idx + 1, k - 1, newtight, sum); # If i is non zero, then k remains # unchanged and value is added to sum else: cnt += solve(idx + 1, k, newtight, sum + i); dp[idx][k][tight][sum] = cnt # Memoize and return return cnt;# Driver codeif __name__=='__main__': # K is the number of exact non-zero # elements to have in number N = 169 k = 2; # break N into its digits ConvertIntoDigit(N); # We keep record of 0s we need to # place in the number k = len(digits) - k; print(solve(0, k, 0, 0))# This code is contributed by rutvik_56. |
C#
// C# program to count the numbers having// exactly K non-zero digits and sum// of digits are odd and distinct.using System;using System.Collections.Generic;class GFG{// To store digits of Nstatic List<int> digits = new List<int>();// visited mapstatic bool []vis = new bool[170];// DP Tablestatic int [,,,]dp = new int[ 19, 19, 2, 170 ];// Push all the digits of N into// digits vectorstatic void ConvertIntoDigit(int n){ while (n > 0) { int dig = n % 10; digits.Add(dig); n /= 10; } digits.Reverse();}// Function returns the countstatic int solve(int idx, int k, int tight, int sum){ // If desired number is formed // whose sum is odd if (idx == digits.Count && k == 0 && sum % 2 == 1) { // If it is not present in map, // mark it as true and return 1 if (!vis[sum]) { vis[sum] = true; return 1; } // Sum is present in map already return 0; } // Desired result not found if (idx > digits.Count) { return 0; } // If that state is already calculated // just return that state value if (dp[idx, k, tight, sum] > 0) { return dp[idx, k, tight, sum]; } // Upper limit int j; if (idx < digits.Count && tight == 0) { j = digits[idx]; } else { j = 9; } // To store the count of // desired numbers int cnt = 0; // If k is non-zero, i ranges from // 0 to j else [1, j] for(int i = (k > 0 ? 0 : 1); i <= j; i++) { int newtight = tight; if (i < j) { newtight = 1; } // If current digit is 0, decrement // k and recurse sum is not changed // as we are just adding 0 that // makes no difference if (i == 0) cnt += solve(idx + 1, k - 1, newtight, sum); // If i is non zero, then k remains // unchanged and value is added to sum else cnt += solve(idx + 1, k, newtight, sum + i); } // Memoize and return return dp[idx, k, tight, sum] = cnt;}// Driver codepublic static void Main(String[] args){ // K is the number of exact non-zero // elements to have in number int N, k; N = 169; k = 2; // break N into its digits ConvertIntoDigit(N); // We keep record of 0s we need to // place in the number k = digits.Count - k; Console.Write(solve(0, k, 0, 0));}}// This code is contributed by amal kumar choubey |
Javascript
<script> // JavaScript program to count the numbers having // exactly K non-zero digits and sum // of digits are odd and distinct. // To store digits of N let digits = []; // visited map let vis = new Array(170); vis.fill(false); // DP Table let dp = new Array(19); for(let i = 0; i < 19; i++) { dp[i] = new Array(19); for(let j = 0; j < 19; j++) { dp[i][j] = new Array(2); for(let k = 0; k < 2; k++) { dp[i][j][k] = new Array(170); for(let l = 0; l < 170; l++) { dp[i][j][k][l] = 0; } } } } // Push all the digits of N into // digits vector function ConvertIntoDigit(n) { while (n > 0) { let dig = n % 10; digits.push(dig); n = parseInt(n / 10, 10); } digits.reverse(); } // Function returns the count function solve(idx, k, tight, sum) { // If desired number is formed // whose sum is odd if (idx == digits.length && k == 0 && sum % 2 == 1) { // If it is not present in map, // mark it as true and return 1 if (!vis[sum]) { vis[sum] = true; return 1; } // Sum is present in map already return 0; } // Desired result not found if (idx > digits.length) { return 0; } // If that state is already calculated // just return that state value if (dp[idx][k][tight][sum] > 0) { return dp[idx][k][tight][sum]; } // Upper limit let j; if (idx < digits.length && tight == 0) { j = digits[idx]; } else { j = 9; } // To store the count of // desired numbers let cnt = 0; // If k is non-zero, i ranges from // 0 to j else [1, j] for(let i = (k > 0 ? 0 : 1); i <= j; i++) { let newtight = tight; if (i < j) { newtight = 1; } // If current digit is 0, decrement // k and recurse sum is not changed // as we are just adding 0 that // makes no difference if (i == 0) cnt += solve(idx + 1, k - 1, newtight, sum); // If i is non zero, then k remains // unchanged and value is added to sum else cnt += solve(idx + 1, k, newtight, sum + i); } // Memoize and return dp[idx][k][tight][sum] = cnt; return dp[idx][k][tight][sum]; } // K is the number of exact non-zero // elements to have in number let N, k; N = 169; k = 2; // break N into its digits ConvertIntoDigit(N); // We keep record of 0s we need to // place in the number k = digits.length - k; document.write(solve(0, k, 0, 0));</script> |
Output:
12
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