Ways to form an array having integers in given range such that total sum is divisible by 2
Last Updated :
07 May, 2025
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
Explanation: 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 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].
Steps to solve the problem:
- Initialize tL to l and tR to r.
- Initialize arrays L and R with 0 for each type (modulo 0 and 1). Set L[l % 2] to l and R[r % 2] to r.
- Increment l and decrement r.
- If l is less than or equal to tR and r is greater than or equal to tL, set L[l % 2] to l and R[r % 2] to r.
- Initialize cnt0 and cnt1 as zero.
- If R[0] and L[0] are non-zero, set cnt0 to (R[0] - L[0]) / 2 + 1.
- If R[1] and L[1] are non-zero, set cnt1 to (R[1] - L[1]) / 2 + 1.
- Initialize a 2D array dp of size n x 2 with all elements set to 0.
- Set dp[1][0] to cnt0 and dp[1][1] to cnt1.
- For i from 2 to n:
- Set dp[i][0] to cnt0 times dp[i - 1][0] plus cnt1 times dp[i - 1][1].
- Set dp[i][1] to cnt0 times dp[i - 1][1] plus cnt1 times dp[i - 1][0].
- Return dp[n][0] as the required count of ways.
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 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 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# 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
<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>
<?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
?>
Time Complexity: O(n)
Auxiliary Space: O(n)
Efficient approach : Space optimization O(1)
In previous approach we the current value dp[i] is only depend upon the previous 2 values i.e. dp[i-1][0] and dp[i-1][1]. So to optimize the space we can keep track of previous and current values by the help of three variables prevRow0, prevRow1 and currRow0 , currRow1 which will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Create 2 variables prevRow0 and prevRow1 to keep track of previous values of DP.
- Initialize base case prevRow0 = cnt0 , prevRow1 = cnt1. where cnt is Count of numbers of each type between range
- Create a variable currRow0 and currRow1 to store current value.
- Iterate over subproblem using loop and update curr.
- After every iteration update prevRow0 and prevRow1 for further iterations.
- At last return prevRow0.
Implementation:
C++
#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;
// Initialize variables for the first row
int prevRow0 = cnt0, prevRow1 = cnt1;
// Iterate through the rows and update the previous row values
for (int i = 2; i <= n; i++) {
int currRow0 = (cnt0 * prevRow0) + (cnt1 * prevRow1);
int currRow1 = (cnt0 * prevRow1) + (cnt1 * prevRow0);
prevRow0 = currRow0;
prevRow1 = currRow1;
}
// Return the required count of ways
return prevRow0;
}
// Driver Code
int main()
{
int n = 2, l = 1, r = 3;
cout << countWays(n, l, r);
return 0;
}
import java.util.*;
public class Main {
// Function to return the number of ways to
// form an array of size n such that sum of
// all elements is divisible by 2
public 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[2];
int[] R = new int[2];
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;
}
// Initialize variables for the first row
int prevRow0 = cnt0, prevRow1 = cnt1;
// Iterate through the rows and update the previous
// row values
for (int i = 2; i <= n; i++) {
int currRow0
= (cnt0 * prevRow0) + (cnt1 * prevRow1);
int currRow1
= (cnt0 * prevRow1) + (cnt1 * prevRow0);
prevRow0 = currRow0;
prevRow1 = currRow1;
}
// Return the required count of ways
return prevRow0;
}
// Driver Code
public static void main(String[] args)
{
int n = 2, l = 1, r = 3;
System.out.println(countWays(n, l, r));
}
}
# 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, R = [0, 0], [0, 0]
L[l % 2] = l
R[r % 2] = r
l += 1
r -= 1
if l <= tR and r >= tL:
L[l % 2] = l
R[r % 2] = 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
# Initialize variables for the first row
prevRow0, prevRow1 = cnt0, cnt1
# Iterate through the rows and update the previous row values
for i in range(2, n+1):
currRow0 = (cnt0 * prevRow0) + (cnt1 * prevRow1)
currRow1 = (cnt0 * prevRow1) + (cnt1 * prevRow0)
prevRow0 = currRow0
prevRow1 = currRow1
# Return the required count of ways
return prevRow0
# Driver Code
if __name__ == '__main__':
n, l, r = 2, 1, 3
print(countWays(n, l, r))
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 = { 0, 0 };
int[] R = { 0, 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] != 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;
// Initialize variables for the first row
int prevRow0 = cnt0, prevRow1 = cnt1;
// Iterate through the rows and update the previous row values
for (int i = 2; i <= n; i++)
{
int currRow0 = (cnt0 * prevRow0) + (cnt1 * prevRow1);
int currRow1 = (cnt0 * prevRow1) + (cnt1 * prevRow0);
prevRow0 = currRow0;
prevRow1 = currRow1;
}
// Return the required count of ways
return prevRow0;
}
// Driver Code
static void Main(string[] args)
{
int n = 2, l = 1, r = 3;
Console.WriteLine(CountWays(n, l, r));
// Wait for key press to exit
Console.ReadLine();
}
}
// 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;
let tR = r;
// Represents first and last numbers of each type (modulo 0 and 1)
let L = [0, 0];
let R = [0, 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
let cnt0 = 0;
let cnt1 = 0;
if (R[0] !== 0 && L[0] !== 0)
cnt0 = Math.floor((R[0] - L[0]) / 2) + 1;
if (R[1] !== 0 && L[1] !== 0)
cnt1 = Math.floor((R[1] - L[1]) / 2) + 1;
// Initialize variables for the first row
let prevRow0 = cnt0;
let prevRow1 = cnt1;
// Iterate through the rows and update the previous row values
for (let i = 2; i <= n; i++) {
let currRow0 = cnt0 * prevRow0 + cnt1 * prevRow1;
let currRow1 = cnt0 * prevRow1 + cnt1 * prevRow0;
prevRow0 = currRow0;
prevRow1 = currRow1;
}
// Return the required count of ways
return prevRow0;
}
// Driver Code
const n = 2;
const l = 1;
const r = 3;
console.log(countWays(n, l, r));
Output:
5
Time Complexity: O(n)
Auxiliary Space: O(1)
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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