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[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++ 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(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 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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