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

• Last Updated : 07 May, 2021

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[0] – 1 to X, now the X is 15. This X is not good.
Then add arr[1] – 1 to X, now the X is 15 + 16 = 31. 31 % H = 7. This X is also not good.
Then add arr[2] to X, now the X is 7 + 14 = 21. This X is good.
Then add arr[3] – 1 to X, now the X is (21 + 19) % H = 16. This X is not good.
Then add arr[4] to X, now the X is (16 + 20) % H = 12. This X is not good.
Then add arr[5] to X, now the X is 12 + 11 = 23. This X is good.
Then add arr[6] 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++

 `// 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][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]``                    ``= 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[0]);` `    ``goodInteger(A, size, H, L, R);` `    ``return` `0;``}`

## Java

 `// 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

 `# 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#

 `// 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`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N * H)

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