Given three positive integers **N**, **L** and **R**. The task is to find the number of ways to form an array of size **N** where each element lies in the range **[L, R]** such that the total sum of all the elements of the array is divisible by **2**.

**Examples:**

Input:N = 2, L = 1, R = 3

Output:5

Possible arrays having sum of all elements divisible by 2 are

[1, 1], [2, 2], [1, 3], [3, 1] and [3, 3]

Input:N = 3, L = 2, R = 2

Output:1

**Approach:** The idea is to find the count of numbers having remainder 0 and 1 modulo 2 separately lying between L and R. This count can be calculated as follows:

We need to count numbers between range having remainder 1 modulo 2

F= First number in range of required type

L= Last number in range of required type

Count = (L – F) / 2

cnt0, and cnt1 represents Count of numbers between range of each type.

Then, using dynamic programming we can solve this problem. Let **dp[i][j]** denotes the number of ways where the sum of first i numbers modulo 2 is equal to j. Suppose we need to calculate dp[i][0], then it will have the following recurrence relation: **dp[i][0] = (cnt0 * dp[i – 1][0] + cnt1 * dp[i – 1][1])**. First term represents the number of ways upto (i – 1) having sum remainder as 0, so we can place cnt0 numbers in i^{th} position such that sum remainder still remains 0. Second term represents the number of ways upto (i – 1) having sum remainder as 1, so we can place cnt1 numbers in i^{th} position to such that sum remainder becomes 0. Similarly, we can calculate for dp[i][1].

Final answer will be denoted by **dp[N][0]**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the number of ways to ` `// form an array of size n such that sum of ` `// all elements is divisible by 2 ` `int` `countWays(` `int` `n, ` `int` `l, ` `int` `r) ` `{ ` ` ` `int` `tL = l, tR = r; ` ` ` ` ` `// Represents first and last numbers ` ` ` `// of each type (modulo 0 and 1) ` ` ` `int` `L[2] = { 0 }, R[2] = { 0 }; ` ` ` `L[l % 2] = l, R[r % 2] = r; ` ` ` ` ` `l++, r--; ` ` ` ` ` `if` `(l <= tR && r >= tL) ` ` ` `L[l % 2] = l, R[r % 2] = r; ` ` ` ` ` `// Count of numbers of each type between range ` ` ` `int` `cnt0 = 0, cnt1 = 0; ` ` ` `if` `(R[0] && L[0]) ` ` ` `cnt0 = (R[0] - L[0]) / 2 + 1; ` ` ` `if` `(R[1] && L[1]) ` ` ` `cnt1 = (R[1] - L[1]) / 2 + 1; ` ` ` ` ` `int` `dp[n][2]; ` ` ` ` ` `// Base Cases ` ` ` `dp[1][0] = cnt0; ` ` ` `dp[1][1] = cnt1; ` ` ` `for` `(` `int` `i = 2; i <= n; i++) { ` ` ` ` ` `// Ways to form array whose sum upto ` ` ` `// i numbers modulo 2 is 0 ` ` ` `dp[i][0] = (cnt0 * dp[i - 1][0] ` ` ` `+ cnt1 * dp[i - 1][1]); ` ` ` ` ` `// Ways to form array whose sum upto ` ` ` `// i numbers modulo 2 is 1 ` ` ` `dp[i][1] = (cnt0 * dp[i - 1][1] ` ` ` `+ cnt1 * dp[i - 1][0]); ` ` ` `} ` ` ` ` ` `// Return the required count of ways ` ` ` `return` `dp[n][0]; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `n = 2, l = 1, r = 3; ` ` ` `cout << countWays(n, l, r); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 implementation of the approach

# Function to return the number of ways to

# form an array of size n such that sum of

# all elements is divisible by 2

def countWays(n, l, r):

tL, tR = l, r

# Represents first and last numbers

# of each type (modulo 0 and 1)

L = [0 for i in range(2)]

R = [0 for i in range(2)]

L[l % 2] = l

R[r % 2] = r

l += 1

r -= 1

if (l <= tR and r >= tL):

L[l % 2], R[r % 2] = l, r

# Count of numbers of each type

# between range

cnt0, cnt1 = 0, 0

if (R[0] and L[0]):

cnt0 = (R[0] – L[0]) // 2 + 1

if (R[1] and L[1]):

cnt1 = (R[1] – L[1]) // 2 + 1

dp = [[0 for i in range(2)]

for i in range(n + 1)]

# Base Cases

dp[1][0] = cnt0

dp[1][1] = cnt1

for i in range(2, n + 1):

# Ways to form array whose sum

# upto i numbers modulo 2 is 0

dp[i][0] = (cnt0 * dp[i – 1][0] +

cnt1 * dp[i – 1][1])

# Ways to form array whose sum upto

# i numbers modulo 2 is 1

dp[i][1] = (cnt0 * dp[i – 1][1] +

cnt1 * dp[i – 1][0])

# Return the required count of ways

return dp[n][0]

# Driver Code

n, l, r = 2, 1, 3

print(countWays(n, l, r))

# This code is contributed

# by Mohit Kumar

**Output:**

5

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