Count Numbers in Range with difference between Sum of digits at even and odd positions as Prime

Given a range [L, R]. The task is to count the numbers in the range having difference between the sum of digits at even position and sum of digits at odd position is a Prime Number. Consider the position of least significant digit in the number as an odd position.
Examples:

Input : L = 1, R = 50
Output : 6
Explanation : Only, 20, 30, 31, 41, 
42 and 50 are valid numbers. 

Input : L = 50, R = 100
Output : 18

Prerequisites: Digit DP
Approach: Firstly, if we are able to count the required numbers up to R i.e. in the range [0, R], we can easily reach our answer in the range [L, R] by solving for from zero to R and then subtracting the answer we get after solving from zero to L – 1. Now, we need to define the DP states.
DP States:

• Since we can consider our number as a sequence of digits, one state is the position at which we are currently at. This position can have values from 0 to 18 if we are dealing with numbers up to 10¹⁸. In each recursive call, we try to build the sequence from left to right by placing a digit from 0 to 9.
• First state is the sum of the digits at even positions we have placed so far.
• Second state is the sum of the digits at odd positions we have placed so far.
• Another state is the boolean variable tight which tells the number we are trying to build has already become smaller than R so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, the maximum limit of digit we can place is digit at the current position in R.

Also, when we reach the base condition, we need to check whether the required difference is a prime number or not. Since the highest number in range is 10¹⁸, the maximum sum at either even or odd positions can be at max 9 times 9 and hence the maximum difference. So, we need to check only prime numbers only upto 100 at base condition.
Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

const int M = 18;
int a, b, dp[M][90][90][2];

// Prime numbers upto 100
int prime[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23,
                29, 31, 37, 43, 47, 53, 59, 61,
                67, 71, 73, 79, 83, 89, 97 };

// Function to return the count of
// required numbers from 0 to num
int count(int pos, int even, int odd, int tight,
        vector<int> num)
{
    // Base Case
    if (pos == num.size()) {
        if (num.size() & 1)
            swap(odd, even);
        int d = even - odd;

        // check if the difference is equal
        // to any prime number
        for (int i = 0; i < 24; i++)
            if (d == prime[i])
                return 1;

        return 0;
    }

    // If this result is already computed
    // simply return it
    if (dp[pos][even][odd][tight] != -1)
        return dp[pos][even][odd][tight];

    int ans = 0;

    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    int limit = (tight ? 9 : num[pos]);

    for (int d = 0; d <= limit; d++) {
        int currF = tight, currEven = even;
        int currOdd = odd;

        if (d < num[pos])
            currF = 1;

        // If the current position is odd
        // add it to currOdd, otherwise to
        // currEven
        if (pos & 1)
            currOdd += d;
        else
            currEven += d;

        ans += count(pos + 1, currEven, currOdd,
                    currF, num);
    }

    return dp[pos][even][odd][tight] = ans;
}

// Function to convert x into its digit vector
// and uses count() function to return the
// required count
int solve(int x)
{
    vector<int> num;

    while (x) {
        num.push_back(x % 10);
        x /= 10;
    }

    reverse(num.begin(), num.end());

    // Initialize dp
    memset(dp, -1, sizeof(dp));
    return count(0, 0, 0, 0, num);
}

// Driver Code
int main()
{
    int L = 1, R = 50;
    cout << solve(R) - solve(L - 1) << endl;

    L = 50, R = 100;
    cout << solve(R) - solve(L - 1) << endl;

    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG
{

static int M = 18;
static int a, b, dp[][][][] = new int[M][90][90][2];

// Prime numbers upto 100
static int prime[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23,
                29, 31, 37, 43, 47, 53, 59, 61,
                67, 71, 73, 79, 83, 89, 97 };

// Function to return the count of
// required numbers from 0 to num
static int count(int pos, int even, int odd, int tight,
        Vector<Integer> num)
{
    // Base Case
    if (pos == num.size())
    {
        if ((num.size() & 1) != 0)
        {
            int t = odd;
            odd = even;
            even = t;

        }
        int d = even - odd;

        // check if the difference is equal
        // to any prime number
        for (int i = 0; i < 24; i++)
            if (d == prime[i])
                return 1;

        return 0;
    }

    // If this result is already computed
    // simply return it
    if (dp[pos][even][odd][tight] != -1)
        return dp[pos][even][odd][tight];

    int ans = 0;

    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    int limit = (tight != 0 ? 9 : num.get(pos));

    for (int d = 0; d <= limit; d++)
    {
        int currF = tight, currEven = even;
        int currOdd = odd;

        if (d < num.get(pos))
            currF = 1;

        // If the current position is odd
        // add it to currOdd, otherwise to
        // currEven
        if ((pos & 1) != 0)
            currOdd += d;
        else
            currEven += d;

        ans += count(pos + 1, currEven, currOdd,
                    currF, num);
    }

    return dp[pos][even][odd][tight] = ans;
}

// Function to convert x into its digit vector
// and uses count() function to return the
// required count
static int solve(int x)
{
    Vector<Integer> num = new Vector<Integer>();

    while (x != 0)
    {
        num.add(x % 10);
        x /= 10;
    }

    Collections.reverse(num);

    // Initialize dp
    for(int i = 0; i < dp.length; i++)
        for(int j = 0; j < dp[i].length; j++)
            for(int k = 0; k < dp[i][j].length; k++)
                for(int k1 = 0; k1 < dp[i][j][k].length; k1++)
                    dp[i][j][k][k1] = -1;

    return count(0, 0, 0, 0, num);
}

// Driver Code
public static void main(String args[])
{
    int L = 1, R = 50;
    System.out.println( solve(R) - solve(L - 1));

    L = 50; R = 100;
    System.out.println( solve(R) - solve(L - 1));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python implementation of the above approach

M = 18

# Prime numbers upto 100
prime = [ 2, 3, 5, 7, 11, 13, 17, 19, 23,
        29, 31, 37, 43, 47, 53, 59, 61,
        67, 71, 73, 79, 83, 89, 97 ]

# Function to return the count of
# required numbers from 0 to num
def count(pos, even, odd, tight, num):

    # Base Case
    if pos == len(num):
        if len(num) & 1:
            odd, even = even, odd

        d = even - odd

        # check if the difference is equal
        # to any prime number
        for i in range(24):
            if d == prime[i]:
                return 1
        return 0

    # If this result is already computed
    # simply return it
    if dp[pos][even][odd][tight] != -1:
        return dp[pos][even][odd][tight]

    ans = 0

    # Maximum limit upto which we can place
    # digit. If tight is 1, means number has
    # already become smaller so we can place
    # any digit, otherwise num[pos]
    limit = 9 if tight else num[pos]

    for d in range(limit + 1):
        currF = tight
        currEven = even
        currOdd = odd

        if d < num[pos]:
            currF = 1

        # If the current position is odd
        # add it to currOdd, otherwise to
        # currEven
        if pos & 1:
            currOdd += d
        else:
            currEven += d

        ans += count(pos + 1, currEven, currOdd, currF, num)

    dp[pos][even][odd][tight] = ans
    return dp[pos][even][odd][tight]

# Function to convert x into its digit vector
# and uses count() function to return the
# required count
def solve(x):
    global dp
    num = []
    while x:
        num.append(x % 10)
        x //= 10

    num.reverse()

    # Initialize dp
    dp = [[[[-1, -1] for i in range(90)]
            for j in range(90)] for k in range(M)]
    return count(0, 0, 0, 0, num)

# Driver Code
if __name__ == "__main__":
    dp = []

    L = 1
    R = 50
    print(solve(R) - solve(L - 1))

    L = 50
    R = 100
    print(solve(R) - solve(L - 1))

# This code is contributed by
# sanjeev2552

C#

// C# implementation of the above approach
using System;
using System.Collections.Generic;

class GFG
{
static int M = 18;
static int a, b;
static int [,,,]dp = new int[M, 90, 90, 2];

// Prime numbers upto 100
static int []prime = { 2, 3, 5, 7, 11, 13, 17, 19, 23,
                       29, 31, 37, 43, 47, 53, 59, 61,
                       67, 71, 73, 79, 83, 89, 97 };

// Function to return the count of
// required numbers from 0 to num
static int count(int pos, int even,
                 int odd, int tight,
                 List<int> num)
{
    // Base Case
    if (pos == num.Count)
    {
        if ((num.Count & 1) != 0)
        {
            int t = odd;
            odd = even;
            even = t;

        }

        int d = even - odd;

        // check if the difference is equal
        // to any prime number
        for (int i = 0; i < 24; i++)
            if (d == prime[i])
                return 1;

        return 0;
    }

    // If this result is already computed
    // simply return it
    if (dp[pos, even, odd, tight] != -1)
        return dp[pos, even, odd, tight];

    int ans = 0;

    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    int limit = (tight != 0 ? 9 : num[pos]);

    for (int d = 0; d <= limit; d++)
    {
        int currF = tight, currEven = even;
        int currOdd = odd;

        if (d < num[pos])
            currF = 1;

        // If the current position is odd
        // add it to currOdd, otherwise to
        // currEven
        if ((pos & 1) != 0)
            currOdd += d;
        else
            currEven += d;

        ans += count(pos + 1, currEven,
                     currOdd, currF, num);
    }

    return dp[pos, even, odd, tight] = ans;
}

// Function to convert x into its digit vector
// and uses count() function to return the
// required count
static int solve(int x)
{
    List<int> num = new List<int>();

    while (x != 0)
    {
        num.Add(x % 10);
        x /= 10;
    }

    num.Reverse();

    // Initialize dp
    for(int i = 0; i < dp.GetLength(0); i++)
        for(int j = 0; j < dp.GetLength(1); j++)
            for(int k = 0; k < dp.GetLength(2); k++)
                for(int k1 = 0; k1 < dp.GetLength(3); k1++)
                    dp[i, j, k, k1] = -1;

    return count(0, 0, 0, 0, num);
}

// Driver Code
public static void Main(String []args)
{
    int L = 1, R = 50;
    Console.WriteLine(solve(R) - solve(L - 1));

    L = 50; R = 100;
    Console.WriteLine(solve(R) - solve(L - 1));
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript implementation of the above approach
var M = 18;
var a, b;
var dp = Array.from(Array(M), ()=>Array(90));

// Prime numbers upto 100
var prime = [2, 3, 5, 7, 11, 13, 17, 19, 23,
                29, 31, 37, 43, 47, 53, 59, 61,
                67, 71, 73, 79, 83, 89, 97 ];

// Function to return the count of
// required numbers from 0 to num
function count(pos, even, odd, tight, num)
{
    // Base Case
    if (pos == num.length) {
        if (num.length & 1)
            [odd, even] = [even, odd]
        var d = even - odd;

        // check if the difference is equal
        // to any prime number
        for (var i = 0; i < 24; i++)
            if (d == prime[i])
                return 1;

        return 0;
    }

    // If this result is already computed
    // simply return it
    if (dp[pos][even][odd][tight] != -1)
        return dp[pos][even][odd][tight];

    var ans = 0;

    // Maximum limit upto which we can place
    // digit. If tight is 1, means number has
    // already become smaller so we can place
    // any digit, otherwise num[pos]
    var limit = (tight ? 9 : num[pos]);

    for (var d = 0; d <= limit; d++) {
        var currF = tight, currEven = even;
        var currOdd = odd;

        if (d < num[pos])
            currF = 1;

        // If the current position is odd
        // add it to currOdd, otherwise to
        // currEven
        if (pos & 1)
            currOdd += d;
        else
            currEven += d;

        ans += count(pos + 1, currEven, currOdd,
                    currF, num);
    }

    return dp[pos][even][odd][tight] = ans;
}

// Function to convert x into its digit vector
// and uses count() function to return the
// required count
function solve(x)
{
    var num = [];

    while (x) {
        num.push(x % 10);
        x = parseInt(x/10);
    }

    num.reverse();

    // Initialize dp
    for(var i =0; i<M; i++)
        for(var j =0; j<90; j++)
            dp[i][j] = Array.from(Array(90),  ()=>Array(2).fill(-1))
    return count(0, 0, 0, 0, num);
}

// Driver Code
var L = 1, R = 50;
document.write( solve(R) - solve(L - 1) + "<br>");
L = 50, R = 100;
document.write( solve(R) - solve(L - 1));

</script>

Output

6
18

Time Complexity: O(pos*limit)

Auxiliary Space: O(M*90*90*2)