Sum of all numbers formed having 4 atmost X times, 5 atmost Y times and 6 atmost Z times

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Given three integers X, Y and Z, the task is to find the sum of all the numbers formed having 4 at most X times, 5 at most Y times, and 6 at most Z times, under mod 10^9+7.
Examples:

Input: X = 1, Y = 1, Z = 1 
Output: 3675
Explanation:
4 + 5 + 6 + 45 + 54 + 56 
+ 65 + 46 + 64 + 456 + 465 
+ 546 + 564 + 645 + 654 = 3675

Input: X = 4, Y = 5, Z = 6
Output: 129422134

Approach:

As this problem has the property of sub-problems overlapping and optimal sub-structure, hence dynamic programming can be used to solve it.
The numbers having exact i 4s, j 5s, and k 6s for all i < x, j < y, j < z are required to get the required sum.
Therefore the DP array exactnum[i][j][k] will store the exact count of numbers having exact i 4s, j 5s, and k 6s.
If exactnum[i – 1][j][k], exactnum[i][j – 1][k] and exactnum[i][j][k – 1] are already known, then it can be observed that the sum of these is the required answer, except in the case when exactnum[i – 1][j][k], exactnum[i][j – 1][k] or exactnum[i][j][k – 1] doesn’t exist. In that case, just skip it.
exactsum[i][j][k] stores the sum of the exact number having i 4’s, j 5’s, and k 6’s in the same way as

exactsum[i][j][k] = 10 * (exactsum[i - 1][j][k] 
                        + exactsum[i][j - 1][k] 
                        + exactsum[i][j][k - 1]) 
                  + 4 * exactnum[i - 1][j][k] 
                  + 5 * exactnum[i][j - 1][k] 
                  + 6 * exactnum[i][j][k - 1]

Below is the implementation of the above approach:

C++

// C++ program to find sum of all numbers
// formed having 4 atmost X times, 5 atmost
// Y times and 6 atmost Z times
#include <bits/stdc++.h>
using namespace std;
 
const int N = 101;
const int mod = 1e9 + 7;
 
// exactsum[i][j][k] stores the sum of
// all the numbers having exact
// i 4's, j 5's and k 6's
int exactsum[N][N][N];
 
// exactnum[i][j][k] stores numbers
// of numbers having exact
// i 4's, j 5's and k 6's
int exactnum[N][N][N];
 
// Utility function to calculate the
// sum for x 4's, y 5's and z 6's
int getSum(int x, int y, int z)
{
    int ans = 0;
    exactnum[0][0][0] = 1;
    for (int i = 0; i <= x; ++i) {
        for (int j = 0; j <= y; ++j) {
            for (int k = 0; k <= z; ++k) {
 
                // Computing exactsum[i][j][k]
                // as explained above
                if (i > 0) {
                    exactsum[i][j][k]
                        += (exactsum[i - 1][j][k] * 10
                            + 4 * exactnum[i - 1][j][k])
                           % mod;
                    exactnum[i][j][k]
                        += exactnum[i - 1][j][k] % mod;
                }
                if (j > 0) {
                    exactsum[i][j][k]
                        += (exactsum[i][j - 1][k] * 10
                            + 5 * exactnum[i][j - 1][k])
                           % mod;
                    exactnum[i][j][k]
                        += exactnum[i][j - 1][k] % mod;
                }
                if (k > 0) {
                    exactsum[i][j][k]
                        += (exactsum[i][j][k - 1] * 10
                            + 6 * exactnum[i][j][k - 1])
                           % mod;
                    exactnum[i][j][k]
                        += exactnum[i][j][k - 1] % mod;
                }
 
                ans += exactsum[i][j][k] % mod;
                ans %= mod;
            }
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int x = 1, y = 1, z = 1;
 
    cout << (getSum(x, y, z) % mod);
 
    return 0;
}

Java

// Java program to find sum of all numbers 
// formed having 4 atmost X times, 5 atmost 
// Y times and 6 atmost Z times 
     
class GFG 
{
     
    static int N = 101; 
    static int mod = (int)1e9 + 7; 
     
    // exactsum[i][j][k] stores the sum of 
    // all the numbers having exact 
    // i 4's, j 5's and k 6's 
    static int exactsum[][][] = new int[N][N][N]; 
     
    // exactnum[i][j][k] stores numbers 
    // of numbers having exact 
    // i 4's, j 5's and k 6's 
    static int exactnum[][][] = new int[N][N][N]; 
     
    // Utility function to calculate the 
    // sum for x 4's, y 5's and z 6's 
    static int getSum(int x, int y, int z) 
    { 
        int ans = 0; 
        exactnum[0][0][0] = 1; 
        for (int i = 0; i <= x; ++i)
        { 
            for (int j = 0; j <= y; ++j) 
            { 
                for (int k = 0; k <= z; ++k) 
                { 
     
                    // Computing exactsum[i][j][k] 
                    // as explained above 
                    if (i > 0)
                    { 
                        exactsum[i][j][k] 
                        += (exactsum[i - 1][j][k] * 10
                        + 4 * exactnum[i - 1][j][k]) % mod; 
                         
                        exactnum[i][j][k] 
                        += exactnum[i - 1][j][k] % mod; 
                    } 
                    if (j > 0)
                    { 
                        exactsum[i][j][k] 
                        += (exactsum[i][j - 1][k] * 10
                        + 5 * exactnum[i][j - 1][k]) % mod; 
                         
                        exactnum[i][j][k] 
                        += exactnum[i][j - 1][k] % mod; 
                    } 
                    if (k > 0)
                    { 
                        exactsum[i][j][k] 
                        += (exactsum[i][j][k - 1] * 10
                        + 6 * exactnum[i][j][k - 1]) % mod; 
                         
                        exactnum[i][j][k] 
                        += exactnum[i][j][k - 1] % mod; 
                    } 
     
                    ans += exactsum[i][j][k] % mod; 
                    ans %= mod; 
                } 
            } 
        } 
        return ans; 
    } 
     
    // Driver code 
    public static void main (String[] args)
    { 
        int x = 1, y = 1, z = 1; 
     
        System.out.println(getSum(x, y, z) % mod); 
     
    } 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to find sum of all numbers 
# formed having 4 atmost X times, 5 atmost 
# Y times and 6 atmost Z times 
import numpy as np
 
N = 101; 
mod = int(1e9) + 7; 
 
# exactsum[i][j][k] stores the sum of 
# all the numbers having exact 
# i 4's, j 5's and k 6's 
exactsum = np.zeros((N, N, N)); 
 
# exactnum[i][j][k] stores numbers 
# of numbers having exact 
# i 4's, j 5's and k 6's 
exactnum = np.zeros((N, N, N)); 
 
# Utility function to calculate the 
# sum for x 4's, y 5's and z 6's 
def getSum(x, y, z) : 
    ans = 0; 
    exactnum[0][0][0] = 1; 
    for i in range(x + 1) :
        for j in range(y + 1) :
            for k in range(z + 1) :
 
                # Computing exactsum[i][j][k] 
                # as explained above 
                if (i > 0) :
                    exactsum[i][j][k] += (exactsum[i - 1][j][k] * 10 +
                                            4 * exactnum[i - 1][j][k]) % mod;
                                             
                    exactnum[i][j][k] += exactnum[i - 1][j][k] % mod; 
                 
                if (j > 0) :
                    exactsum[i][j][k] += (exactsum[i][j - 1][k] * 10+
                                        5 * exactnum[i][j - 1][k]) % mod; 
                                         
                    exactnum[i][j][k] += exactnum[i][j - 1][k] % mod; 
                 
                if (k > 0) :
                    exactsum[i][j][k] += (exactsum[i][j][k - 1] * 10
                                            + 6 * exactnum[i][j][k - 1]) % mod; 
                    exactnum[i][j][k] += exactnum[i][j][k - 1] % mod; 
 
                ans += exactsum[i][j][k] % mod; 
                ans %= mod; 
                 
    return ans; 
 
# Driver code 
if __name__ == "__main__" : 
 
    x = 1; y = 1; z = 1; 
 
    print((getSum(x, y, z) % mod)); 
 
# This code is contributed by AnkitRai01

C#

// C# program to find sum of all numbers 
// formed having 4 atmost X times, 5 atmost 
// Y times and 6 atmost Z times 
using System;
 
class GFG 
{
     
    static int N = 101; 
    static int mod = (int)1e9 + 7; 
     
    // exactsum[i][j][k] stores the sum of 
    // all the numbers having exact 
    // i 4's, j 5's and k 6's 
    static int [,,]exactsum = new int[N, N, N]; 
     
    // exactnum[i][j][k] stores numbers 
    // of numbers having exact 
    // i 4's, j 5's and k 6's 
    static int [,,]exactnum= new int[N, N, N]; 
     
    // Utility function to calculate the 
    // sum for x 4's, y 5's and z 6's 
    static int getSum(int x, int y, int z) 
    { 
        int ans = 0; 
        exactnum[0, 0, 0] = 1; 
        for (int i = 0; i <= x; ++i)
        { 
            for (int j = 0; j <= y; ++j) 
            { 
                for (int k = 0; k <= z; ++k) 
                { 
     
                    // Computing exactsum[i, j, k] 
                    // as explained above 
                    if (i > 0)
                    { 
                        exactsum[i, j, k] 
                        += (exactsum[i - 1, j, k] * 10 
                        + 4 * exactnum[i - 1, j, k]) % mod; 
                         
                        exactnum[i, j, k] 
                        += exactnum[i - 1, j, k] % mod; 
                    } 
                    if (j > 0)
                    { 
                        exactsum[i, j, k] 
                        += (exactsum[i, j - 1, k] * 10 
                        + 5 * exactnum[i, j - 1, k]) % mod; 
                         
                        exactnum[i, j, k] 
                        += exactnum[i, j - 1, k] % mod; 
                    } 
                    if (k > 0)
                    { 
                        exactsum[i, j, k] 
                        += (exactsum[i, j, k - 1] * 10 
                        + 6 * exactnum[i, j, k - 1]) % mod; 
                         
                        exactnum[i, j, k] 
                        += exactnum[i, j, k - 1] % mod; 
                    } 
     
                    ans += exactsum[i, j, k] % mod; 
                    ans %= mod; 
                } 
            } 
        } 
        return ans; 
    } 
     
    // Driver code 
    public static void Main ()
    { 
        int x = 1, y = 1, z = 1; 
     
        Console.WriteLine(getSum(x, y, z) % mod); 
     
    } 
}
     
// This code is contributed by AnkitRai01

Javascript

<script>
    // Javascript program to find sum of all numbers 
    // formed having 4 atmost X times, 5 atmost 
    // Y times and 6 atmost Z times
     
    let N = 101; 
    let mod = 1e9 + 7; 
       
    // exactsum[i][j][k] stores the sum of 
    // all the numbers having exact 
    // i 4's, j 5's and k 6's 
    let exactsum = new Array(N); 
       
    // exactnum[i][j][k] stores numbers 
    // of numbers having exact 
    // i 4's, j 5's and k 6's 
    let exactnum = new Array(N);
     
    for(let i = 0; i < N; i++)
    {
        exactsum[i] = new Array(N);
        exactnum[i] = new Array(N);
        for(let j = 0; j < N; j++)
        {
            exactsum[i][j] = new Array(N);
            exactnum[i][j] = new Array(N);
            for(let k = 0; k < N; k++)
            {
                exactsum[i][j][k] = 0;
                exactnum[i][j][k] = 0;
            }
        }
    }
       
    // Utility function to calculate the 
    // sum for x 4's, y 5's and z 6's 
    function getSum(x, y, z) 
    { 
        let ans = 0; 
        exactnum[0][0][0] = 1; 
        for (let i = 0; i <= x; ++i)
        { 
            for (let j = 0; j <= y; ++j) 
            { 
                for (let k = 0; k <= z; ++k) 
                { 
       
                    // Computing exactsum[i][j][k] 
                    // as explained above 
                    if (i > 0)
                    { 
                        exactsum[i][j][k] 
                        += (exactsum[i - 1][j][k] * 10 
                        + 4 * exactnum[i - 1][j][k]) % mod; 
                           
                        exactnum[i][j][k] 
                        += exactnum[i - 1][j][k] % mod; 
                    } 
                    if (j > 0)
                    { 
                        exactsum[i][j][k] 
                        += (exactsum[i][j - 1][k] * 10 
                        + 5 * exactnum[i][j - 1][k]) % mod; 
                           
                        exactnum[i][j][k] 
                        += exactnum[i][j - 1][k] % mod; 
                    } 
                    if (k > 0)
                    { 
                        exactsum[i][j][k] 
                        += (exactsum[i][j][k - 1] * 10 
                        + 6 * exactnum[i][j][k - 1]) % mod; 
                           
                        exactnum[i][j][k] 
                        += exactnum[i][j][k - 1] % mod; 
                    } 
       
                    ans += exactsum[i][j][k] % mod; 
                    ans %= mod; 
                } 
            } 
        } 
        return ans; 
    } 
     
    let x = 1, y = 1, z = 1; 
       
    document.write(getSum(x, y, z) % mod);
 
</script>

Output:

Time Complexity: O(x*y*z)

Auxiliary Space: O(N³)

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