Count of Numbers in Range where the number does not contain more than K non zero digits

Given a range represented by two positive integers L and R and a positive integer K. Find the count of numbers in the range where the number does not contain more than K non zero digits.

Examples:

Input : L = 1, R = 1000, K = 3
Output : 1000
Explanation : All the numbers from 1 to 1000 
are 3 digit numbers which obviously cannot 
contain more than 3 non zero digits.

Input : L = 9995, R = 10005
Output : 6
Explanation : Required numbers are 
10000, 10001, 10002, 10003, 10004 and 10005

Prerequisites : Digit DP



There can be two approaches to solve this type of problem, one can be a combinatorial solution and other can be a dynamic programming based solution. Below is a detailed approach of solving this problem using a digit dynamic programming approach.
Dynamic Programming Solution : Firstly, if we are able to count the required numbers upto R i.e. in the range [0, R], we can easily reach our answer in the range [L, R] by solving for from zero to R and then subtracting the answer we get after solving for from zero to L – 1. Now, we need to define the DP states.
DP States:

  • Since we can consider our number as a sequence of digits, one state is the position at which we are currently in. This position can have values from 0 to 18 if we are dealing with the numbers upto 1018. In each recursive call, we try to build the sequence from left to right by placing a digit from 0 to 9.
  • Second state is the count which defines the number of non zero digits, we have placed in the number we are trying to build.
  • Another state is the boolean variable tight which tells the number we are trying to build has already become smaller than R, so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, maximum limit of digit we can place is digit at current position in R.

In the final recursive call, when we are at the last position if the count of non zero digits is less than or equal to K, return 1 otherwise return 0.

Below is the implementation of the above approach.

C++

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// CPP Program to find the count of
// numbers in a range where the number
// does not contain more than K non
// zero digits
  
#include <bits/stdc++.h>
using namespace std;
  
const int M = 20;
  
// states - position, count, tight
int dp[M][M][2];
  
// K is the number of non zero digits
int K;
  
// This function returns the count of
// required numbers from 0 to num
int countInRangeUtil(int pos, int cnt, int tight,
                     vector<int> num)
{
    // Last position
    if (pos == num.size()) {
        // If count of non zero digits
        // is less than or equal to K
        if (cnt <= K)
            return 1;
        return 0;
    }
  
    // If this result is already computed
    // simply return it
    if (dp[pos][cnt][tight] != -1)
        return dp[pos][cnt][tight];
  
    int ans = 0;
  
    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    int limit = (tight ? 9 : num[pos]);
  
    for (int dig = 0; dig <= limit; dig++) {
        int currCnt = cnt;
  
        // If the current digit is nonzero
        // increment currCnt
        if (dig != 0)
            currCnt++;
  
        int currTight = tight;
  
        // At this position, number becomes
        // smaller
        if (dig < num[pos])
            currTight = 1;
  
        // Next recursive call
        ans += countInRangeUtil(pos + 1, currCnt,
                                currTight, num);
    }
    return dp[pos][cnt][tight] = ans;
}
  
// This function converts a number into its
// digit vector and uses above function to compute
// the answer
int countInRange(int x)
{
    vector<int> num;
    while (x) {
        num.push_back(x % 10);
        x /= 10;
    }
    reverse(num.begin(), num.end());
  
    // Initialize dp
    memset(dp, -1, sizeof(dp));
    return countInRangeUtil(0, 0, 0, num);
}
  
// Driver Code to test above functions
int main()
{
    int L = 1, R = 1000;
    K = 3;
    cout << countInRange(R) - countInRange(L - 1) << endl;
  
    L = 9995, R = 10005, K = 2;
    cout << countInRange(R) - countInRange(L - 1) << endl;
    return 0;
}

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Java

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// Java Program to find the count of 
// numbers in a range where the number 
// does not contain more than K non 
// zero digits
import java.util.*;
class Solution
{
static final int M = 20
  
// states - position, count, tight 
static int dp[][][]= new int[M][M][2]; 
  
// K is the number of non zero digits 
static int K; 
static Vector<Integer> num;
  
// This function returns the count of 
// required numbers from 0 to num 
static int countInRangeUtil(int pos, int cnt, int tight ) 
    // Last position 
    if (pos == num.size()) { 
        // If count of non zero digits 
        // is less than or equal to K 
        if (cnt <= K) 
            return 1
        return 0
    
  
    // If this result is already computed 
    // simply return it 
    if (dp[pos][cnt][tight] != -1
        return dp[pos][cnt][tight]; 
  
    int ans = 0
  
    // Maximum limit upto which we can place 
    // digit. If tight is 1, means number has 
    // already become smaller so we can place 
    // any digit, otherwise num[pos] 
    int limit = (tight!=0 ? 9 : num.get(pos)); 
  
    for (int dig = 0; dig <= limit; dig++) { 
        int currCnt = cnt; 
  
        // If the current digit is nonzero 
        // increment currCnt 
        if (dig != 0
            currCnt++; 
  
        int currTight = tight; 
  
        // At this position, number becomes 
        // smaller 
        if (dig < num.get(pos)) 
            currTight = 1
  
        // Next recursive call 
        ans += countInRangeUtil(pos + 1, currCnt, currTight); 
    
    return dp[pos][cnt][tight] = ans; 
  
// This function converts a number into its 
// digit vector and uses above function to compute 
// the answer 
static int countInRange(int x) 
    num= new Vector<Integer>(); 
    while (x!=0) { 
        num.add(x % 10); 
        x /= 10
    
    Collections.reverse(num); 
  
    // Initialize dp 
    for(int i=0;i<M;i++)
        for(int j=0;j<M;j++)
            for(int k=0;k<2;k++)
            dp[i][j][k]=-1;
    return countInRangeUtil(0, 0, 0); 
  
// Driver Code to test above functions 
public static void main(String args[])
    int L = 1, R = 1000
    K = 3
    System.out.println( countInRange(R) - countInRange(L - 1) ); 
  
    L = 9995; R = 10005; K = 2
    System.out.println(  countInRange(R) - countInRange(L - 1) ); 
}
//contributed by Arnab Kundu

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Output:

1000
6

Time Complexity : O(18 * 18 * 2 * 10), if we are dealing with the numbers upto 1018



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