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

Given two numbers start and end as range variables. The goal is to find the count of numbers that lie in this range [start,end] and have a difference of sum of digits at even and sum of digits at odd positions as Prime.

That is (sum of digits at even position)-(sum of digits at odd position) = a Prime number

## Let us understand with examples.

**For Example**

**Input -** start = 230, end = 270

**Output -** Count of Numbers in Range with difference between Sum of digits at even and odd positions as Prime are: 6

**Explanation -** The number(s) between 230 to 270 that meet the condition are:

240 ( 4-2 is 2 ), 250 ( 5-2 is 3 ), 251 ( 5-3 is 2 ), 261 ( 6-3 is 3 ), 262 ( 6-4 is 2 ), 270 ( 7-2 is 5 ).

All these differences are 2, 3 and 5 which are primes.

**Input -** start = 1101, end = 1120

**Output -** Count of Numbers in Range with difference between Sum of digits at even and odd positions as Prime are: 1

**Explanation -** The number(s) between 1101 to 1120 that meet the condition are:

1120 ( 3-1 is 2 ). 2 is prime.

## Approach used in the below program is as follows

In this we use a dynamic programming approach and store the counts of numbers that have Prime differences of sum of even and odd position digits. This array would be arr[size][90][90][2]. Here size is the power of 10. So the largest number as input will be 10^{size}.

In each recursive call to function check(int place, int eve, int od, int temp, vector<int> vec) we will build a number by placing digits 0 to 9 from left to right.

In arr[size][x][y][temp], x is for sums of digits at even positions placed upto x and y is for sums of odd digits placed upto y. Check if the required difference is Prime or not using array arr_2[] which stores all prime numbers upto 100 in order.

- Take variables start and end as input.
- Take global array arr[size][90][90][2] and array arr_2[] for primes up to 100.
- Function check(int place, int eve, int od, int temp, vector<int> vec) takes current position of digit as place, current sum of eve position digits as even and odd position digits as od, value of temp and vector vec which has digits.
- It populates values at arr[place][eve][od][temp] recursively.
- Take initial value for current element as count=0.
- For current position, check if place is last position using if(place == vec.size()). If yes check if that position is odd or even.
- if(vec.size() & 1) results true then current position is odd so swap eve with od as it is odd length number.
- Calculate temp_2 as difference of sums as eve-od.
- Using for loop, traverse arr_2[] and check if temp_2 is found. If yes then its prime. So return 1 else return 0.
- If arr[place][eve][od][temp] is already computed then it would not be -1 so return it.
- If temp is non-zero then set temp_3=9. Temp_3 is the maximum limit of the digit which we can place. If it is 0 then place vec[place] otherwise the number is already smaller so place any digit say 9.
- Traverse digits from 0 to temp_3. If the current position is odd then update set_odd = set_odd + i; ( previous odd position sum + current digit i ).
- If the current position is even then update set_even = set_even + i; ( previous even position sum + current digit i ).
- Set count += check(place + 1, set_even, set_odd, set_temp, vec); and return arr[place][eve][od][temp] = count.
- Function place_prime(int val) takes the number val and generates a vector vec containing its digit from LSB to MSB.
- Set the whole array arr[][][][] with -1.
- Take count = check(0, 0, 0, 0, vec) which will return the result at end.
- Return the count as result.

## Example

#include <bits/stdc++.h> using namespace std; const int size = 18; int arr[size][90][90][2]; //firt 100 prime Numbers int arr_2[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 }; int check(int place, int eve, int od, int temp, vector < int > vec) { int count; int temp_3; if (place == vec.size()) { if (vec.size() & 1) { swap(od, eve); } int temp_2 = eve - od; for (int i = 0; i < 24; i++) { if (temp_2 == arr_2[i]) { return 1; } } return 0; } if (arr[place][eve][od][temp] != -1) { int set = arr[place][eve][od][temp]; return set; } if (temp) { temp_3 = 9; } else { temp_3 = vec[place]; } for (int i = 0; i <= temp_3; i++) { int set_temp = temp; int set_even = eve; int set_odd = od; if (i < vec[place]) { set_temp = 1; } if (place & 1) { set_odd = set_odd + i; } else { set_even = set_even + i; } count += check(place + 1, set_even, set_odd, set_temp, vec); } return arr[place][eve][od][temp] = count; } int place_prime(int val) { vector < int > vec; while (val) { vec.push_back(val % 10); val = val / 10; } reverse(vec.begin(), vec.end()); memset(arr, -1, sizeof(arr)); int count = check(0, 0, 0, 0, vec); return count; } int main() { int start = 20, end = 80; int count = place_prime(end) - place_prime(start - 1); cout << "Count of Numbers in Range with difference between Sum of digits at even and odd positions as Prime are: " << count; return 0; }

If we run the above code it will generate the following output −

## Output

Count of Numbers in Range with difference between Sum of digits at even and odd positions as Prime are: 15

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