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

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```

Recommended Practice

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 ` `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

 ``

Output:

`3675`

Time Complexity: O(x*y*z)

Auxiliary Space: O(N3)

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