title: Compute Ncr P Set 1 Introduction And Dynamic Programming Solution
signature: computeNcrPSet1IntroductionAndDynamicProgrammingSolution(n, r, p) {
test_code: import 'dart:convert';
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  bool validateSolution(int actual, int expected){
      return actual.toString() == expected.toString();
  }
  
  String driver(n, r, p, int expected){
      try
      {
          if (validateSolution(PLACEHOLDER_FN_NAME(n, r, p), expected)){
              return "PASSED";
          }
          return "FAILED";
      }catch(exception_object){
          return exception_object.runtimeType.toString();
      }
  }
  
  int main() {
  
      print("TEST-0...${driver(
          82, 5, 94, 
          50)
      }");
  
      print("TEST-1...${driver(
          45, 24, 95, 
          80)
      }");
  
      print("TEST-2...${driver(
          44, 68, 61, 
          0)
      }");
  
      print("TEST-3...${driver(
          88, 24, 43, 
          0)
      }");
  
      print("TEST-4...${driver(
          90, 75, 57, 
          0)
      }");
  
      print("TEST-5...${driver(
          98, 55, 43, 
          2)
      }");
  
      print("TEST-6...${driver(
          80, 54, 88, 
          0)
      }");
  
      print("TEST-7...${driver(
          60, 75, 65, 
          0)
      }");
  
      print("TEST-8...${driver(
          52, 73, 86, 
          0)
      }");
  
      print("TEST-9...${driver(
          71, 26, 45, 
          12)
      }");
  
      return 0;
  }
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 ];
  }