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Count of non decreasing Arrays with ith element in range [A[i], B[i]]

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  • Last Updated : 28 Oct, 2021
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Given two arrays A[] and B[] both consisting of N integers, the task is to find the number of non-decreasing arrays of size N that can be formed such that each array element lies over the range [A[i], B[i]].

Examples:

Input: A[] = {1, 1}, B[] = {2, 3}
Output: 5
Explanation:
The total number of valid arrays are {1, 1}, {1, 2}, {1, 3}, {2, 2}, {2, 3}. Therefore, the count of such arrays is 5.

Input: A[] = {3, 4, 5}, B[] = {4, 5, 6}
Output: 8

 

Approach: The given problem can be solved using Dynamic Programming and Prefix Sum. Follow the steps below to solve the problem:

  • Initialize a 2D array dp[][] with values 0, where dp[i][j] denotes the total valid array till position i and with the current element as j. Initialize dp[0][0] as 1.
  • Initialize a 2D array pref[][] with values 0 to store the prefix sum of the array.
  • Iterate over the range [0, B[N – 1]] using the variable i and set the value of pref[0][i] as 1.
  • Iterate over the range [1, N] using the variable i and perform the following steps:
    • Iterate over the range [A[i – 1], B[i – 1]] using the variable j and increment the value of dp[i][j] by pref[i – 1][j] and increase the value of pref[i][j] by dp[i][j].
    • Iterate over the range [0, B[N – 1]] using the variable j and if j is greater than 0 then update the prefix sum table by incrementing the value of pref[i][j] by pref[i][j – 1].
  • Initialize the variable ans as 0 to store the resultant count of arrays formed.
  • Iterate over the range [A[N – 1], B[N – 1]] using the variable i and add the value of dp[N][i] to the variable ans.
  • After performing the above steps, print the value of ans as the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the total number
// of possible valid arrays
int totalValidArrays(int a[], int b[],
                     int N)
{
    // Make a 2D DP table
    int dp[N + 1][b[N - 1] + 1];
 
    // Make a 2D prefix sum table
    int pref[N + 1][b[N - 1] + 1];
 
    // Initialize all values to 0
    memset(dp, 0, sizeof(dp)),
        memset(pref, 0, sizeof(pref));
 
    // Base Case
    dp[0][0] = 1;
 
    // Initialize the prefix values
    for (int i = 0; i <= b[N - 1]; i++) {
        pref[0][i] = 1;
    }
 
    // Iterate over the range and update
    // the dp table accordingly
    for (int i = 1; i <= N; i++) {
        for (int j = a[i - 1];
             j <= b[i - 1]; j++) {
            dp[i][j] += pref[i - 1][j];
 
            // Add the dp values to the
            // prefix sum
            pref[i][j] += dp[i][j];
        }
 
        // Update the prefix sum table
        for (int j = 0; j <= b[N - 1]; j++) {
            if (j > 0) {
                pref[i][j] += pref[i][j - 1];
            }
        }
    }
 
    // Find the result count of
    // arrays formed
    int ans = 0;
    for (int i = a[N - 1];
         i <= b[N - 1]; i++) {
        ans += dp[N][i];
    }
 
    // Return the total count of arrays
    return ans;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 1 };
    int B[] = { 2, 3 };
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << totalValidArrays(A, B, N);
 
    return 0;
}

Java




// Java program for the above approach
public class GFG {
     
    // Function to count the total number
    // of possible valid arrays
    static int totalValidArrays(int a[], int b[],
                         int N)
    {
        // Make a 2D DP table
        int dp[][] = new int[N + 1][b[N - 1] + 1];
     
        // Make a 2D prefix sum table
        int pref[][] = new int[N + 1][b[N - 1] + 1];
     
        // Initialize all values to 0
        for (int i = 0; i < N + 1; i++)
            for (int j = 0; j < b[N - 1] + 1; j++)
                dp[i][j] = 0;
                 
        for (int i = 0; i < N + 1; i++)
            for (int j = 0; j < b[N - 1] + 1; j++)
                pref[i][j] = 0;       
 
        // Base Case
        dp[0][0] = 1;
     
        // Initialize the prefix values
        for (int i = 0; i <= b[N - 1]; i++) {
            pref[0][i] = 1;
        }
     
        // Iterate over the range and update
        // the dp table accordingly
        for (int i = 1; i <= N; i++) {
            for (int j = a[i - 1];
                 j <= b[i - 1]; j++) {
                dp[i][j] += pref[i - 1][j];
     
                // Add the dp values to the
                // prefix sum
                pref[i][j] += dp[i][j];
            }
     
            // Update the prefix sum table
            for (int j = 0; j <= b[N - 1]; j++) {
                if (j > 0) {
                    pref[i][j] += pref[i][j - 1];
                }
            }
        }
     
        // Find the result count of
        // arrays formed
        int ans = 0;
        for (int i = a[N - 1];
             i <= b[N - 1]; i++) {
            ans += dp[N][i];
        }
     
        // Return the total count of arrays
        return ans;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int A[] = { 1, 1 };
        int B[] = { 2, 3 };
        int N = A.length;
     
        System.out.println(totalValidArrays(A, B, N));
    }
}
 
// This code is contributed by AnkThon

Python3




# python program for the above approach
 
# Function to count the total number
# of possible valid arrays
def totalValidArrays(a, b, N):
 
    # Make a 2D DP table
    dp = [[0 for _ in range(b[N - 1] + 1)] for _ in range(N + 1)]
 
    # Make a 2D prefix sum table
    pref = [[0 for _ in range(b[N - 1] + 1)] for _ in range(N + 1)]
 
    # Base Case
    dp[0][0] = 1
 
    # Initialize the prefix values
    for i in range(0, b[N - 1] + 1):
        pref[0][i] = 1
 
    # Iterate over the range and update
    # the dp table accordingly
    for i in range(1, N + 1):
        for j in range(a[i - 1], b[i - 1] + 1):
            dp[i][j] += pref[i - 1][j]
 
            # Add the dp values to the
            # prefix sum
            pref[i][j] += dp[i][j]
 
        # Update the prefix sum table
        for j in range(0, b[N - 1] + 1):
            if (j > 0):
                pref[i][j] += pref[i][j - 1]
 
    # Find the result count of
    # arrays formed
    ans = 0
    for i in range(a[N - 1], b[N - 1] + 1):
        ans += dp[N][i]
 
    # Return the total count of arrays
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    A = [1, 1]
    B = [2, 3]
    N = len(A)
 
    print(totalValidArrays(A, B, N))
 
    # This code is contributed by rakeshsahni

C#




// C#  program for the above approach
using System;
class GFG
{
 
  // Function to count the total number
  // of possible valid arrays
  static int totalValidArrays(int[] a, int[] b, int N)
  {
    // Make a 2D DP table
    int[,] dp = new int[N + 1, b[N - 1] + 1];
 
    // Make a 2D prefix sum table
    int[,] pref = new int[N + 1, b[N - 1] + 1];
 
    // Initialize all values to 0
    for (int i = 0; i < N + 1; i++)
      for (int j = 0; j < b[N - 1] + 1; j++)
        dp[i, j] = 0;
 
    for (int i = 0; i < N + 1; i++)
      for (int j = 0; j < b[N - 1] + 1; j++)
        pref[i, j] = 0;       
 
    // Base Case
    dp[0, 0] = 1;
 
    // Initialize the prefix values
    for (int i = 0; i <= b[N - 1]; i++) {
      pref[0, i] = 1;
    }
 
    // Iterate over the range and update
    // the dp table accordingly
    for (int i = 1; i <= N; i++) {
      for (int j = a[i - 1];
           j <= b[i - 1]; j++) {
        dp[i, j] += pref[i - 1, j];
 
        // Add the dp values to the
        // prefix sum
        pref[i, j] += dp[i, j];
      }
 
      // Update the prefix sum table
      for (int j = 0; j <= b[N - 1]; j++) {
        if (j > 0) {
          pref[i, j] += pref[i, j - 1];
        }
      }
    }
 
    // Find the result count of
    // arrays formed
    int ans = 0;
    for (int i = a[N - 1];
         i <= b[N - 1]; i++) {
      ans += dp[N, i];
    }
 
    // Return the total count of arrays
    return ans;
  }
 
  // Driver Code
  public static void Main ()
  {  
    int[] A = { 1, 1 };
    int[] B = { 2, 3 };
    int N = A.Length;
 
    Console.WriteLine(totalValidArrays(A, B, N));
  }
}
 
// This code is contributed by Saurabh

Javascript




<script>
    // JavaScript program for the above approach
 
    // Function to count the total number
    // of possible valid arrays
    const totalValidArrays = (a, b, N) => {
        // Make a 2D DP table
        let dp = new Array(N + 1).fill(0).map(() => new Array(b[N - 1] + 1).fill(0));
 
        // Make a 2D prefix sum table
        let pref = new Array(N + 1).fill(0).map(() => new Array(b[N - 1] + 1).fill(0));
 
        // Base Case
        dp[0][0] = 1;
 
        // Initialize the prefix values
        for (let i = 0; i <= b[N - 1]; i++) {
            pref[0][i] = 1;
        }
 
        // Iterate over the range and update
        // the dp table accordingly
        for (let i = 1; i <= N; i++) {
            for (let j = a[i - 1];
                j <= b[i - 1]; j++) {
                dp[i][j] += pref[i - 1][j];
 
                // Add the dp values to the
                // prefix sum
                pref[i][j] += dp[i][j];
            }
 
            // Update the prefix sum table
            for (let j = 0; j <= b[N - 1]; j++) {
                if (j > 0) {
                    pref[i][j] += pref[i][j - 1];
                }
            }
        }
 
        // Find the result count of
        // arrays formed
        let ans = 0;
        for (let i = a[N - 1];
            i <= b[N - 1]; i++) {
            ans += dp[N][i];
        }
 
        // Return the total count of arrays
        return ans;
    }
 
    // Driver Code
    let A = [1, 1];
    let B = [2, 3];
    let N = A.length;
 
    document.write(totalValidArrays(A, B, N));
 
// This code is contributed by rakeshsahni
 
</script>

Output: 

5

 

Time Complexity: O(N*M), where M is the last element of the array B[].
Auxiliary Space: O(N*M), where M is the last element of the array B[].


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