# Maximize occurrences of values between L and R on sequential addition of Array elements with modulo H

Given an array arr[] having N positive integers and a positive integers H, the task is to count the maximum occurrences of a value from the range [L, R] on adding arr[i] or arr[i] – 1 to X (initially 0). The integer X must be always be less than H. If it is greater than H, replace X by X % H
Examples:

Input: arr = {16, 17, 14, 20, 20, 11, 22}, H = 24, L = 21, R = 23
Output:
Explanation:
Initially X = 0.
Then add arr – 1 to X, now the X is 15. This X is not good.
Then add arr – 1 to X, now the X is 15 + 16 = 31. 31 % H = 7. This X is also not good.
Then add arr to X, now the X is 7 + 14 = 21. This X is good.
Then add arr – 1 to X, now the X is (21 + 19) % H = 16. This X is not good.
Then add arr to X, now the X is (16 + 20) % H = 12. This X is not good.
Then add arr to X, now the X is 12 + 11 = 23. This X is good.
Then add arr to X, now the X is 23 + 22 = 21. This X is also good.
So, the maximum number of good X in the example is 3.
Input: arr = {1, 2, 3}, H = 5, L = 1, R = 2
Output:

Approach: This problem can be solved with dynamic programming. Maintain a table dp[N+1][H] which represents the maximum occurrences of an element in the range [L, R] on adding upto i elements. For every ith index, calculate the maximum possible frequency obtainable by adding arr[i] and arr[i] – 1. Once, computed for all indices, find the maximum from the last row of the dp[][] matrix.
Below is the implementation of above approach:

 `// C++ implementation of the ` `// above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that prints the number ` `// of times X gets a value ` `// between L and R ` `void` `goodInteger(``int` `arr[], ``int` `n, ` `                 ``int` `h, ``int` `l, ``int` `r) ` `{ ` ` `  `    ``vector > dp( ` `        ``n + 1, ` `        ``vector<``int``>(h, -1)); ` ` `  `    ``// Base condition ` `    ``dp = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = 0; j < h; j++) { ` ` `  `            ``// Condition if X can be made ` `            ``// equal to j after i additions ` `            ``if` `(dp[i][j] != -1) { ` ` `  `                ``// Compute value of X ` `                ``// after adding arr[i] ` `                ``int` `h1 = (j + arr[i]) % h; ` ` `  `                ``// Compute value of X after ` `                ``// adding arr[i] - 1 ` `                ``int` `h2 = (j + arr[i] - 1) % h; ` ` `  `                ``// Update dp as the maximum value ` `                ``dp[i + 1][h1] ` `                    ``= max(dp[i + 1][h1], ` `                          ``dp[i][j] ` `                              ``+ (h1 >= l ` `                                 ``&& h1 <= r)); ` `                ``dp[i + 1][h2] ` `                    ``= max(dp[i + 1][h2], ` `                          ``dp[i][j] ` `                              ``+ (h2 >= l ` `                                 ``&& h2 <= r)); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``int` `ans = 0; ` ` `  `    ``// Compute maximum answer from all ` `    ``// possible cases ` `    ``for` `(``int` `i = 0; i < h; i++) { ` `        ``if` `(dp[n][i] != -1) ` `            ``ans = max(ans, dp[n][i]); ` `    ``} ` ` `  `    ``// Printing maximum good occurrence of X ` `    ``cout << ans << ``"\n"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 16, 17, 14, 20, 20, 11, 22 }; ` `    ``int` `H = 24; ` `    ``int` `L = 21; ` `    ``int` `R = 23; ` ` `  `    ``int` `size = ``sizeof``(A) / ``sizeof``(A); ` ` `  `    ``goodInteger(A, size, H, L, R); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the ` `// above approach ` `class` `GFG{ ` ` `  `// Function that prints the number ` `// of times X gets a value ` `// between L and R ` `static` `void` `goodInteger(``int` `arr[], ``int` `n, ` `                        ``int` `h, ``int` `l, ``int` `r) ` `{ ` `    ``int` `[][]dp = ``new` `int``[n + ``1``][h]; ` `    ``for``(``int` `i = ``0``; i < n; i++)  ` `        ``for``(``int` `j = ``0``; j < h; j++)  ` `            ``dp[i][j] = -``1``; ` `             `  `    ``// Base condition ` `    ``dp[``0``][``0``] = ``0``; ` ` `  `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``for``(``int` `j = ``0``; j < h; j++) ` `        ``{ ` ` `  `            ``// Condition if X can be made ` `            ``// equal to j after i additions ` `            ``if` `(dp[i][j] != -``1``) ` `            ``{ ` ` `  `                ``// Compute value of X ` `                ``// after adding arr[i] ` `                ``int` `h1 = (j + arr[i]) % h; ` ` `  `                ``// Compute value of X after ` `                ``// adding arr[i] - 1 ` `                ``int` `h2 = (j + arr[i] - ``1``) % h; ` ` `  `                ``// Update dp as the maximum value ` `                ``dp[i + ``1``][h1] = Math.max(dp[i + ``1``][h1], ` `                                         ``dp[i][j] +  ` `                                        ``((h1 >= l && ` `                                          ``h1 <= r) ? ` `                                           ``1` `: ``0``)); ` `                ``dp[i + ``1``][h2] = Math.max(dp[i + ``1``][h2], ` `                                         ``dp[i][j] + ` `                                        ``((h2 >= l && ` `                                          ``h2 <= r) ?  ` `                                           ``1` `: ``0``)); ` `            ``} ` `        ``} ` `    ``} ` `    ``int` `ans = ``0``; ` ` `  `    ``// Compute maximum answer from all ` `    ``// possible cases ` `    ``for``(``int` `i = ``0``; i < h; i++) ` `    ``{ ` `        ``if` `(dp[n][i] != -``1``) ` `            ``ans = Math.max(ans, dp[n][i]); ` `    ``} ` ` `  `    ``// Printing maximum good occurrence of X ` `    ``System.out.print(ans + ``"\n"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `A[] = { ``16``, ``17``, ``14``, ``20``, ``20``, ``11``, ``22` `}; ` `    ``int` `H = ``24``; ` `    ``int` `L = ``21``; ` `    ``int` `R = ``23``; ` ` `  `    ``int` `size = A.length; ` ` `  `    ``goodInteger(A, size, H, L, R); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

 `# Python3 implementation of the above approach ` ` `  `# Function that prints the number ` `# of times X gets a value ` `# between L and R ` `def` `goodInteger(arr, n, h, l, r): ` `     `  `    ``dp ``=` `[[``-``1` `for` `i ``in` `range``(h)] ` `              ``for` `j ``in` `range``(n ``+` `1``)] ` ` `  `    ``# Base condition ` `    ``dp[``0``][``0``] ``=` `0` ` `  `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(h): ` ` `  `            ``# Condition if X can be made ` `            ``# equal to j after i additions ` `            ``if``(dp[i][j] !``=` `-``1``): ` ` `  `                ``# Compute value of X ` `                ``# after adding arr[i] ` `                ``h1 ``=` `(j ``+` `arr[i]) ``%` `h ` ` `  `                ``# Compute value of X after ` `                ``# adding arr[i] - 1 ` `                ``h2 ``=` `(j ``+` `arr[i] ``-` `1``) ``%` `h ` ` `  `                ``# Update dp as the maximum value ` `                ``dp[i ``+` `1``][h1] ``=` `max``(dp[i ``+` `1``][h1], ` `                                    ``dp[i][j] ``+`  `                                    ``(h1 >``=` `l ``and` `h1 <``=` `r)) ` ` `  `                ``dp[i ``+` `1``][h2] ``=` `max``(dp[i ``+` `1``][h2], ` `                                    ``dp[i][j] ``+`  `                                    ``(h2 >``=` `l ``and` `h2 <``=` `r)) ` `    ``ans ``=` `0` ` `  `    ``# Compute maximum answer from all ` `    ``# possible cases ` `    ``for` `i ``in` `range``(h): ` `        ``if``(dp[n][i] !``=` `-``1``): ` `            ``ans ``=` `max``(ans, dp[n][i]) ` ` `  `    ``# Printing maximum good occurrence of X ` `    ``print``(ans) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``A ``=` `[ ``16``, ``17``, ``14``, ``20``, ``20``, ``11``, ``22` `] ` `    ``H ``=` `24` `    ``L ``=` `21` `    ``R ``=` `23` ` `  `    ``size ``=` `len``(A) ` `    ``goodInteger(A, size, H, L, R) ` ` `  `# This code is contributed by Shivam Singh `

 `// C# implementation of the  ` `// above approach  ` `using` `System; ` ` `  `class` `GFG{  ` ` `  `// Function that prints the number  ` `// of times X gets a value  ` `// between L and R  ` `static` `void` `goodint(``int` `[]arr, ``int` `n,  ` `                    ``int` `h, ``int` `l, ``int` `r)  ` `{  ` `    ``int` `[,]dp = ``new` `int``[n + 1, h];  ` ` `  `    ``for``(``int` `i = 0; i < n; i++)  ` `        ``for``(``int` `j = 0; j < h; j++)  ` `            ``dp[i, j] = -1;  ` `             `  `    ``// Base condition  ` `    ``dp[0, 0] = 0;  ` ` `  `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{  ` `        ``for``(``int` `j = 0; j < h; j++)  ` `        ``{  ` `             `  `            ``// Condition if X can be made  ` `            ``// equal to j after i additions  ` `            ``if` `(dp[i, j] != -1)  ` `            ``{  ` ` `  `                ``// Compute value of X  ` `                ``// after adding arr[i]  ` `                ``int` `h1 = (j + arr[i]) % h;  ` ` `  `                ``// Compute value of X after  ` `                ``// adding arr[i] - 1  ` `                ``int` `h2 = (j + arr[i] - 1) % h;  ` ` `  `                ``// Update dp as the maximum value  ` `                ``dp[i + 1, h1] = Math.Max(dp[i + 1, h1],  ` `                                         ``dp[i, j] +  ` `                                        ``((h1 >= l &&  ` `                                          ``h1 <= r) ?  ` `                                           ``1 : 0));  ` `                ``dp[i + 1, h2] = Math.Max(dp[i + 1, h2],  ` `                                         ``dp[i, j] +  ` `                                        ``((h2 >= l &&  ` `                                          ``h2 <= r) ?  ` `                                           ``1 : 0));  ` `            ``}  ` `        ``}  ` `    ``}  ` `    ``int` `ans = 0;  ` ` `  `    ``// Compute maximum answer from all  ` `    ``// possible cases  ` `    ``for``(``int` `i = 0; i < h; i++)  ` `    ``{  ` `        ``if` `(dp[n, i] != -1)  ` `            ``ans = Math.Max(ans, dp[n, i]);  ` `    ``}  ` ` `  `    ``// Printing maximum good occurrence of X  ` `    ``Console.Write(ans + ``"\n"``);  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``int` `[]A = { 16, 17, 14, 20, 20, 11, 22 };  ` `    ``int` `H = 24;  ` `    ``int` `L = 21;  ` `    ``int` `R = 23;  ` ` `  `    ``int` `size = A.Length;  ` ` `  `    ``goodint(A, size, H, L, R);  ` `}  ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

Output:
```3

```

Time Complexity: O(N * H)

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