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Ways to form an array having integers in given range such that total sum is divisible by 2

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  • Difficulty Level : Medium
  • Last Updated : 22 Jun, 2022

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:
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:
 

 

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 ith 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 ith 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;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the number of ways to
// form an array of size n such that sum of
// all elements is divisible by 2
static 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 = new int[3];
    int[] R = new int[3];
    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] > 0 && L[0] > 0)
        cnt0 = (R[0] - L[0]) / 2 + 1;
    if (R[1] > 0 && L[1] > 0)
        cnt1 = (R[1] - L[1]) / 2 + 1;
 
    int[][] dp = new int[n + 1][3];
 
    // 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
public static void main(String[] args)
{
    int n = 2, l = 1, r = 3;
    System.out.println(countWays(n, l, r));
}
}
 
// This code is contributed by Code_Mech.

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

C#




// C# implementation of the approach
 
using System;
 
class GFG
{
     
// Function to return the number of ways to
// form an array of size n such that sum of
// all elements is divisible by 2
static 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 = new int[3];
    int[] R = new int[3];
    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] > 0 && L[0] > 0)
        cnt0 = (R[0] - L[0]) / 2 + 1;
    if (R[1] > 0 && L[1] > 0)
        cnt1 = (R[1] - L[1]) / 2 + 1;
 
    int[,] dp=new int[n + 1, 3];
 
    // 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
static void Main()
{
    int n = 2, l = 1, r = 3;
    Console.WriteLine(countWays(n, l, r));
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP 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
function countWays($n, $l, $r)
{
    $tL = $l;
    $tR = $r;
     
    $L = array_fill(0, 2, 0);
    $R = array_fill(0, 2, 0);
     
    // Represents first and last numbers
    // of each type (modulo 0 and 1)
    $L[$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
    $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;
 
    $dp = array();
 
    // Base Cases
    $dp[1][0] = $cnt0;
    $dp[1][1] = $cnt1;
    for ($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
$n = 2;
$l = 1;
$r = 3;
 
echo countWays($n, $l, $r);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
 
    // JavaScript 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
    function countWays(n, l, r)
    {
        let tL = l, tR = r;
 
        // Represents first and last numbers
        // of each type (modulo 0 and 1)
        let L = new Array(3);
        let R = new Array(3);
        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
        let cnt0 = 0, cnt1 = 0;
        if (R[0] > 0 && L[0] > 0)
            cnt0 = (R[0] - L[0]) / 2 + 1;
        if (R[1] > 0 && L[1] > 0)
            cnt1 = (R[1] - L[1]) / 2 + 1;
 
        let dp = new Array(n + 1);
        for (let i = 0; i <= n; i++)
        {   
            dp[i] = new Array(3);
            for (let j = 0; j < 3; j++)
            {
                dp[i][j] = 0;
            }
        }
 
        // Base Cases
        dp[1][0] = cnt0;
        dp[1][1] = cnt1;
        for (let 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];
    }
     
    let n = 2, l = 1, r = 3;
    document.write(countWays(n, l, r));
     
</script>

Output: 

5

 

Time Complexity: O(n)

Auxiliary Space: O(n)


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