Given two integers L and R, the task is to find the count of numbers in the range [L, R] having prime digits at prime positions and non-prime digits at non-prime positions.
Examples:
Input: L = 5, R = 22
Output: 7
Explanation: The numbers 6, 8, 9, 12, 13, 15, and 17 have prime digits at prime positions and non-prime digits at non-prime positions.Input: L = 20, R = 29
Output: 0
Explanation: There are no numbers which have prime digits at prime positions and non-prime digits at non-prime positions.
Naive Approach: The simplest approach to solve the problem is to iterate over the range [L, R]. For every ith number check if the digits of the number is prime at prime positions and non-prime at non-prime positions or not. If found to be true, then increment the count. Finally, print the count obtained.
Time Complexity: O(R – L + 1) * sqrt(R) * log10(R)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Digit DP. Following are the recurrence relation between the dynamic programming states:
If i is a prime at prime digits or non-prime at non-prime digits, then x = 1
pos: Stores position of digits
prime: Check if prime digits are present at prime positions and non-prime digits at non-prime positions are present or not.
st: check if a number contains any leading 0.
end: Maximum possible digits at current position
Follow the steps below to solve the problem:
- Initialize a 4D array, say dp[pos][st][tight][prime].
- Compute the value of dp[pos][st][tight][prime] for the number R using memorization, say cntR.
- Compute the value of dp[pos][st][tight][prime] for the number L – 1 using memorization, say cntL.
- Finally, print the value of (cntR – cntL).
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Store digits of a numbervector<long long int> num;// Store overlapping subproblemslong long int dp[19][2][2][19];// Function to check if a// number is prime or notbool isPrime(long long int n){ // If n is less than // or equal to 1 if (n <= 1) return false; // If n is less than // or equal to 3 if (n <= 3) return true; // If n is a multiple of 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over the range [5, n] for (long long int i = 5; i * i <= n; i = i + 6) { // If n is a multiple of i or (i + 2) if (n % i == 0 || n % (i + 2) == 0) return false; } return true;}// Function to count the required// numbers from the given rangelong long cntNum(long long pos, long long st, long long tight, long long prime){ // Base Case if (pos == num.size()) return 1; // If the subproblems already computed if (dp[pos][st][tight][prime] != -1) return dp[pos][st][tight][prime]; long long int res = 0; // Stores maximum possible // at current digits long long end = (tight == 0) ? num[pos] : 9; // Iterate over all possible digits // at current position for (long long i = 0; i <= end; i++) { // Check if i is the maximum possible // digit at current position or not long long ntight = (i < end) ? 1 : tight; // Check if a number contains // leading 0s or not long long int nzero = (i != 0) ? 1 : st; // If number has only leading zeros // and digit is non-zero if ((nzero == 1) && isPrime(i) && isPrime(prime)) { // Prime digits at prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) { // Non-prime digits at // non-prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } // If the number has only leading zeros // and i is zero, if (nzero == 0) res += cntNum(pos + 1, nzero, ntight, prime); } return dp[pos][st][tight][prime] = res;}// Function to find count of numbers in// range [0, b] whose digits are prime// at prime and non-prime at non-prime poslong long int cntZeroRange(long long int b){ num.clear(); // Insert digits of a number, b while (b > 0) { num.push_back(b % 10); b /= 10; } // Reversing the digits in num reverse(num.begin(), num.end()); // Initializing dp with -1 memset(dp, -1, sizeof(dp)); long long int res = cntNum(0, 0, 0, 1); // Returning the value return res;}// Driver Codeint main(){ // Given range, [L, R] long long int L = 5, R = 22; // Function Call long long int res = cntZeroRange(R) - cntZeroRange(L - 1); // Print answer cout << res << endl; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Store digits of a numberstatic Vector<Integer> num = new Vector<>();// Store overlapping subproblemsstatic int [][][][]dp = new int[19][2][2][19];// Function to check if a// number is prime or notstatic boolean isPrime(int n){ // If n is less than // or equal to 1 if (n <= 1) return false; // If n is less than // or equal to 3 if (n <= 3) return true; // If n is a multiple of 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over the range [5, n] for (int i = 5; i * i <= n; i = i + 6) { // If n is a multiple of i or (i + 2) if (n % i == 0 || n % (i + 2) == 0) return false; } return true;}// Function to count the required// numbers from the given rangestatic int cntNum(int pos, int st, int tight, int prime){ // Base Case if (pos == num.size()) return 1; // If the subproblems already computed if (dp[pos][st][tight][prime] != -1) return dp[pos][st][tight][prime]; int res = 0; // Stores maximum possible // at current digits int end = (tight == 0) ? num.get(pos) : 9; // Iterate over all possible digits // at current position for (int i = 0; i <= end; i++) { // Check if i is the maximum possible // digit at current position or not int ntight = (i < end) ? 1 : tight; // Check if a number contains // leading 0s or not int nzero = (i != 0) ? 1 : st; // If number has only leading zeros // and digit is non-zero if ((nzero == 1) && isPrime(i) && isPrime(prime)) { // Prime digits at prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) { // Non-prime digits at // non-prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } // If the number has only leading zeros // and i is zero, if (nzero == 0) res += cntNum(pos + 1, nzero, ntight, prime); } return dp[pos][st][tight][prime] = res;}// Function to find count of numbers in// range [0, b] whose digits are prime// at prime and non-prime at non-prime posstatic int cntZeroRange(int b){ num.clear(); // Insert digits of a number, b while (b > 0) { num.add(b % 10); b /= 10; } // Reversing the digits in num Collections.reverse(num); // Initializing dp with -1 for (int i = 0; i < 19; i++) for (int j = 0; j < 2; j++) for (int k = 0; k < 2; k++) for (int l = 0; l < 19; l++) dp[i][j][k][l] = -1; int res = cntNum(0, 0, 0, 1); // Returning the value return res;}// Driver Codepublic static void main(String[] args){ // Given range, [L, R] int L = 5, R = 22; // Function Call int res = cntZeroRange(R) - cntZeroRange(L - 1); // Print answer System.out.print(res +"\n");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approachfrom math import ceil, sqrt# Function to check if a# number is prime or notdef isPrime(n): # If n is less than # or equal to 1 if (n <= 1): return False # If n is less than # or equal to 3 if (n <= 3): return True # If n is a multiple of 2 or 3 if (n % 2 == 0 or n % 3 == 0): return False # Iterate over the range [5, n] for i in range(5, ceil(sqrt(n)), 6): # If n is a multiple of i or (i + 2) if (n % i == 0 or n % (i + 2) == 0): return False return True# Function to count the required# numbers from the given rangedef cntNum(pos, st, tight, prime): global dp, num if (pos == len(num)): return 1 # If the subproblems already computed if (dp[pos][st][tight][prime] != -1): return dp[pos][st][tight][prime] res = 0 # Stores maximum possible # at current digits end = num[pos] if (tight == 0) else 9 # Iterate over all possible digits # at current position for i in range(end + 1): # Check if i is the maximum possible # digit at current position or not ntight = 1 if (i < end) else tight # Check if a number contains # leading 0s or not nzero = 1 if (i != 0) else st # If number has only leading zeros # and digit is non-zero if ((nzero == 1) and isPrime(i) and isPrime(prime)): # Prime digits at prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1) if ((nzero == 1) and isPrime(i) == False and isPrime(prime) == False): # Non-prime digits at # non-prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1) # If the number has only leading zeros # and i is zero, if (nzero == 0): res += cntNum(pos + 1, nzero, ntight, prime) dp[pos][st][tight][prime] = res return dp[pos][st][tight][prime]# Function to find count of numbers in# range [0, b] whose digits are prime# at prime and non-prime at non-prime posdef cntZeroRange(b): global num, dp num.clear() while (b > 0): num.append(b % 10) b //= 10 # Reversing the digits in num num = num[::-1] # print(num) dp = [[[[-1 for i in range(19)] for i in range(2)] for i in range(2)] for i in range(19)] res = cntNum(0, 0, 0, 1) # Returning the value return res# Driver Codeif __name__ == '__main__': dp = [[[[-1 for i in range(19)] for i in range(2)] for i in range(2)] for i in range(19)] L, R, num = 5, 22, [] # Function Call res = cntZeroRange(R) - cntZeroRange(L - 1) # Print answer print(res)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Store digits of a number static List<int> num = new List<int>(); // Store overlapping subproblems static int[, , , ] dp = new int[19, 2, 2, 19]; // Function to check if a // number is prime or not static bool isPrime(int n) { // If n is less than // or equal to 1 if (n <= 1) return false; // If n is less than // or equal to 3 if (n <= 3) return true; // If n is a multiple of 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over the range [5, n] for (int i = 5; i * i <= n; i = i + 6) { // If n is a multiple of i or (i + 2) if (n % i == 0 || n % (i + 2) == 0) return false; } return true; } // Function to count the required // numbers from the given range static int cntNum(int pos, int st, int tight, int prime) { // Base Case if (pos == num.Count) return 1; // If the subproblems already computed if (dp[pos, st, tight, prime] != -1) return dp[pos, st, tight, prime]; int res = 0; // Stores maximum possible // at current digits int end = (tight == 0) ? num[pos] : 9; // Iterate over all possible digits // at current position for (int i = 0; i <= end; i++) { // Check if i is the maximum possible // digit at current position or not int ntight = (i < end) ? 1 : tight; // Check if a number contains // leading 0s or not int nzero = (i != 0) ? 1 : st; // If number has only leading zeros // and digit is non-zero if ((nzero == 1) && isPrime(i) && isPrime(prime)) { // Prime digits at prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) { // Non-prime digits at // non-prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } // If the number has only leading zeros // and i is zero, if (nzero == 0) res += cntNum(pos + 1, nzero, ntight, prime); } return dp[pos, st, tight, prime] = res; } // Function to find count of numbers in // range [0, b] whose digits are prime // at prime and non-prime at non-prime pos static int cntZeroRange(int b) { num.Clear(); // Insert digits of a number, b while (b > 0) { num.Add(b % 10); b /= 10; } // Reversing the digits in num num.Reverse(); // Initializing dp with -1 for (int i = 0; i < 19; i++) for (int j = 0; j < 2; j++) for (int k = 0; k < 2; k++) for (int l = 0; l < 19; l++) dp[i, j, k, l] = -1; int res = cntNum(0, 0, 0, 1); // Returning the value return res; } // Driver Code public static void Main(string[] args) { // Given range, [L, R] int L = 5, R = 22; // Function Call int res = cntZeroRange(R) - cntZeroRange(L - 1); // Print answer Console.WriteLine(res + "\n"); }}// This code is contributed by chitranayal. |
Javascript
<script>// JavaScript program for the above approach// Store digits of a numberlet num = [];// Store overlapping subproblemslet dp = new Array(19);// Function to check if a// number is prime or notfunction isPrime(n){ // If n is less than // or equal to 1 if (n <= 1) return false; // If n is less than // or equal to 3 if (n <= 3) return true; // If n is a multiple of 2 or 3 if (n % 2 == 0 || n % 3 == 0) return false; // Iterate over the range [5, n] for (let i = 5; i * i <= n; i = i + 6) { // If n is a multiple of i or (i + 2) if (n % i == 0 || n % (i + 2) == 0) return false; } return true;}// Function to count the required// numbers from the given rangefunction cntNum(pos,st,tight,prime){ // Base Case if (pos == num.length) return 1; // If the subproblems already computed if (dp[pos][st][tight][prime] != -1) return dp[pos][st][tight][prime]; let res = 0; // Stores maximum possible // at current digits let end = (tight == 0) ? num[pos] : 9; // Iterate over all possible digits // at current position for (let i = 0; i <= end; i++) { // Check if i is the maximum possible // digit at current position or not let ntight = (i < end) ? 1 : tight; // Check if a number contains // leading 0s or not let nzero = (i != 0) ? 1 : st; // If number has only leading zeros // and digit is non-zero if ((nzero == 1) && isPrime(i) && isPrime(prime)) { // Prime digits at prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } if ((nzero == 1) && !isPrime(i) && !isPrime(prime)) { // Non-prime digits at // non-prime positions res += cntNum(pos + 1, nzero, ntight, prime + 1); } // If the number has only leading zeros // and i is zero, if (nzero == 0) res += cntNum(pos + 1, nzero, ntight, prime); } return dp[pos][st][tight][prime] = res;}// Function to find count of numbers in// range [0, b] whose digits are prime// at prime and non-prime at non-prime posfunction cntZeroRange(b){ num=[]; // Insert digits of a number, b while (b > 0) { num.push(b % 10); b = Math.floor(b/10); } // Reversing the digits in num num.reverse(); for(let i=0;i<19;i++){ dp[i]=new Array(2); for(let j=0;j<2;j++) { dp[i][j]=new Array(2); for(let k=0;k<2;k++) { dp[i][j][k]=new Array(19); for(let l=0;l<19;l++) { dp[i][j][k][l]=-1; } } }} let res = cntNum(0, 0, 0, 1); // Returning the value return res;}// Driver Code// Given range, [L, R]let L = 5, R = 22;// Function Calllet res= cntZeroRange(R) - cntZeroRange(L - 1);// Print answerdocument.write(res +"<br>");// This code is contributed by avanitrachhadiya2155</script> |
Output:
7
Time Complexity: O(log10(R) * log10(L) sqrt(log10(R))* 10 * 4))
Auxiliary Space: O(log10(R) * log10(L) * 4)
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