title: Compute Ncr P Set 1 Introduction And Dynamic Programming Solution
signature: pub fn compute_ncr_p_set_1_introduction_and_dynamic_programming_solution(n: i32, r: i32, p: i32) -> i32 {
test_code: use std::collections::HashMap;
  use std::collections::HashSet;
  use std::cmp::*;
  use std::collections::*;
  use std::fmt::*; 
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  pub fn validate_solution(actual: i32, expected: i32)->bool{
  return actual==expected
  }
  pub fn driver(n: i32, r: i32, p: i32, expected: i32)->String{
      if validate_solution(PLACEHOLDER_FN_NAME(n, r, p), expected) {
        return "PASSED".to_string();
      };
      return "FAILED".to_string();
  
  }
  
  fn main() {
      println!("TEST-0...{}", driver(
        82, 5, 94, 
        50
      ));
      
      println!("TEST-1...{}", driver(
        45, 24, 95, 
        80
      ));
      
      println!("TEST-2...{}", driver(
        44, 68, 61, 
        0
      ));
      
      println!("TEST-3...{}", driver(
        88, 24, 43, 
        0
      ));
      
      println!("TEST-4...{}", driver(
        90, 75, 57, 
        0
      ));
      
      println!("TEST-5...{}", driver(
        98, 55, 43, 
        2
      ));
      
      println!("TEST-6...{}", driver(
        80, 54, 88, 
        0
      ));
      
      println!("TEST-7...{}", driver(
        60, 75, 65, 
        0
      ));
      
      println!("TEST-8...{}", driver(
        52, 73, 86, 
        0
      ));
      
      println!("TEST-9...{}", driver(
        71, 26, 45, 
        12
      ));
      
  }
source_py: def compute_ncr_p_set_1_introduction_and_dynamic_programming_solution(n, r, p):
      C = [0 for i in range((r + 1))]
      C[0] = 1
      for i in range(1, (n + 1)):
          for j in range(min(i, r), 0, (- 1)):
              C[j] = ((C[j] + C[(j - 1)]) % p)
      return C[r]
source_cpp: int compute_ncr_p_set_1_introduction_and_dynamic_programming_solution(int n, int r, int p) {
    int C [ r + 1 ];
    memset ( C, 0, sizeof ( C ) );
    C [ 0 ] = 1;
    for ( int i = 1;
    i <= n;
    i ++ ) {
      for ( int j = min ( i, r );
      j > 0;
      j -- ) C [ j ] = ( C [ j ] + C [ j - 1 ] ) % p;
    }
    return C [ r ];
  }