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

• Difficulty Level : Medium
• Last Updated : 10 Aug, 2021

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++

 // CPP Program to find the count of// numbers in a range where the number// does not contain more than K non// zero digits #include using namespace std; const int M = 20; // states - position, count, tightint dp[M][M][2]; // K is the number of non zero digitsint K; // This function returns the count of// required numbers from 0 to numint countInRangeUtil(int pos, int cnt, int tight,                     vector 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 answerint countInRange(int x){    vector 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 functionsint 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;}

## Java

 // Java Program to find the count of// numbers in a range where the number// does not contain more than K non// zero digitsimport java.util.*;class Solution{static final int M = 20; // states - position, count, tightstatic int dp[][][]= new int[M][M][2]; // K is the number of non zero digitsstatic int K;static Vector num; // This function returns the count of// required numbers from 0 to numstatic 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 answerstatic int countInRange(int x){    num= new Vector();    while (x!=0) {        num.add(x % 10);        x /= 10;    }    Collections.reverse(num);     // Initialize dp    for(int i=0;i

## Python3

 # Python Program to find the count of# numbers in a range where the number# does not contain more than K non# zero digits # This function returns the count of# required numbers from 0 to numdef countInRangeUtil(pos, cnt, tight, num):     # Last position    if pos == len(num):         # 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]     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]    limit = 9 if tight else num[pos]     for dig in range(limit + 1):        currCnt = cnt         # If the current digit is nonzero        # increment currCnt        if dig != 0:            currCnt += 1         currTight = tight         # At this position, number becomes        # smaller        if dig < num[pos]:            currTight = 1         # Next recursive call        ans += countInRangeUtil(pos + 1, currCnt, currTight, num)     dp[pos][cnt][tight] = ans    return dp[pos][cnt][tight] # This function converts a number into its# digit vector and uses above function to compute# the answerdef countInRange(x):    global dp, K, M     num = []    while x:        num.append(x % 10)        x //= 10     num.reverse()     # Initialize dp    dp = [[[-1, -1] for i in range(M)] for j in range(M)]    return countInRangeUtil(0, 0, 0, num) # Driver Codeif __name__ == "__main__":     # states - position, count, tight    dp = []    M = 20     # K is the number of non zero digits    K = 0     L = 1    R = 1000    K = 3    print(countInRange(R) - countInRange(L - 1))     L = 9995    R = 10005    K = 2    print(countInRange(R) - countInRange(L - 1)) # This code is contributed by# sanjeev2552

## C#

 // C# Program to find the count of// numbers in a range where the number// does not contain more than K non// zero digitsusing System;using System.Collections.Generic; class GFG{     static int M = 20; // states - position, count, tightstatic int [,,]dp = new int[M, M, 2]; // K is the number of non zero digitsstatic int K;static List num; // This function returns the count of// required numbers from 0 to numstatic int countInRangeUtil(int pos,                int cnt, int tight ){    // Last position    if (pos == num.Count)    {        // 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[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);    }    return dp[pos,cnt,tight] = ans;} // This function converts a number into its// digit vector and uses above function to compute// the answerstatic int countInRange(int x){    num = new List();    while (x != 0)    {        num.Add(x % 10);        x /= 10;    }    num.Reverse();     // 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 Codepublic static void Main(){    int L = 1, R = 1000;    K = 3;    Console.WriteLine( countInRange(R) - countInRange(L - 1) );     L = 9995; R = 10005; K = 2;    Console.WriteLine( countInRange(R) - countInRange(L - 1) );}} /* This code contributed by PrinciRaj1992 */

## Javascript



Output:

1000
6

Time Complexity : O(M * M * 2 * 10) where M = log(MAX), here MAX is the maximum number.
Auxiliary Space: O(M*M)

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