Given three integers a, b and c. You need to select one integer from the range [a, b] and one integer from the range [b, c] and add them. The task to calculate the number of ways to obtain the sum for all the numbers in the range [1, b+c].
Examples:
Input: a = 1, b = 2, c = 2
Output: 0, 0, 1, 1
Explanation:
The numbers to be obtained are [1, b+c] = [1, 4] = {1, 2, 3, 4}
Therefore, the number of ways to obtain each are:
1 – can’t be obtained
2 – can’t be obtained
3 – only one way. select {1} from range [a, b] and {2} from range [b, c] – 1 + 2 = 3
4 – only one way. select {2} from range [a, b] and {2} from range [b, c] – 2 + 2 = 4
Input: a = 1, b = 3, c = 4
Output: 0, 0, 0, 1, 2, 2, 1
Simple Approach:
- A simple brute force solution will be to use a nested loop where exterior loop traverses from i = a to i = b and inner loop from j = b to j = c inclusive.
- We will initialise array a of size b + c + 1 with zero. Now in loops, we will increment the index at i+j, i.e., (a[i+j]++).
- We will simply print the array at the end.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void CountWays(int a, int b, int c)
{
int x = b + c + 1;
int arr[x] = { 0 };
for (int i = a; i <= b; i++) {
for (int j = b; j <= c; j++) {
arr[i + j]++;
}
}
for (int i = 1; i < x; i++) {
cout << arr[i] << " ";
}
cout << endl;
}
int main()
{
int a = 1;
int b = 2;
int c = 2;
CountWays(a, b, c);
return 0;
}
|
Java
import java.io.*;
public class GFG{
public static void CountWays(int a, int b,
int c)
{
int x = b + c + 1;
int[] arr = new int[x];
for(int i = a; i <= b; i++)
{
for(int j = b; j <= c; j++)
{
arr[i + j]++;
}
}
for(int i = 1; i < x; i++)
{
System.out.print(arr[i] + " ");
}
}
public static void main(String[] args)
{
int a = 1;
int b = 2;
int c = 2;
CountWays(a, b, c);
}
}
|
Python3
def CountWays(a, b, c):
x = b + c + 1;
arr = [0] * x;
for i in range(a, b + 1):
for j in range(b, c + 1):
arr[i + j] += 1;
for i in range(1, x):
print(arr[i], end = " ");
if __name__ == '__main__':
a = 1;
b = 2;
c = 2;
CountWays(a, b, c);
|
C#
using System;
class GFG{
public static void CountWays(int a, int b,
int c)
{
int x = b + c + 1;
int[] arr = new int[x];
for(int i = a; i <= b; i++)
{
for(int j = b; j <= c; j++)
{
arr[i + j]++;
}
}
for(int i = 1; i < x; i++)
{
Console.Write(arr[i] + " ");
}
}
public static void Main()
{
int a = 1;
int b = 2;
int c = 2;
CountWays(a, b, c);
}
}
|
Javascript
<script>
function CountWays(a, b, c)
{
let x = b + c + 1;
let arr = new Array(x);
arr.fill(0);
for (let i = a; i <= b; i++) {
for (let j = b; j <= c; j++) {
arr[i + j]++;
}
}
for (let i = 1; i < x; i++) {
document.write(arr[i] + " ");
}
document.write("</br>");
}
let a = 1;
let b = 2;
let c = 2;
CountWays(a, b, c);
</script>
|
Time Complexity: O((b-a)*(c-b)), which in the worst case is O(c2)
Auxiliary Space: O(x), as We are using extra space.
Efficient Approach: The idea is to use Prefix Sum logic to solve this problem.
- We will traverse i from [a, b] and for every i we will simply increment the value of starting interval arr[i + b] by 1 and decrement the value of ending interval arr[i + c + 1] by 1.
- Now all we need to do is to calculate the prefix sum of the array ( arr[i]+ = arr[i-1] ) and print the array.
Lets see the approach with the help of an example.
Why does this work?
For example: a = 1, b = 2, c = 2, we will encounter only two values of i
i = 1 = > arr[1+2]++; arr[1+2+1]–;
i = 2 = > arr[2+2]++; arr[2+2+1]–;
arr = {0, 0, 0, 1, 0, -1};
prefix sums:
arr = {0, 0, 0, 1, 1, 0};
Now carefully look and realise that this is our answer.
So what we do at particular index i is arr[i+b]++ and arr[i+c+1]–;
Now we are using prefix sums so all the values will be incremented by 1 between i+b and infinite(We won’t go there but will result in prefix sum increment by 1 and as soon as we do a decrement at i+c+1 all the values between i+c+1 and infinite will be decreased by one.
So effectively all the values in the range [i+b, i+c] are incremented by one, and rest all the values will remain unaffected.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void CountWays(int a, int b, int c)
{
int x = b + c + 2;
int arr[x] = { 0 };
for (int i = 1; i <= b; i++) {
arr[i + b]++;
arr[i + c + 1]--;
}
for (int i = 1; i < x - 1; i++) {
arr[i] += arr[i - 1];
cout << arr[i] << " ";
}
cout << endl;
}
int main()
{
int a = 1;
int b = 2;
int c = 2;
CountWays(a, b, c);
return 0;
}
|
Java
import java.io.*;
class GFG{
static void CountWays(int a, int b, int c)
{
int x = b + c + 2;
int arr[] = new int[x];
for(int i = 1; i <= b; i++)
{
arr[i + b]++;
arr[i + c + 1]--;
}
for(int i = 1; i < x - 1; i++)
{
arr[i] += arr[i - 1];
System.out.print(arr[i] + " ");
}
System.out.println();
}
public static void main(String[] args)
{
int a = 1;
int b = 2;
int c = 2;
CountWays(a, b, c);
}
}
|
Python3
def CountWays(a, b, c):
x = b + c + 2;
arr = [0] * x;
for i in range(1, b+1):
arr[i + b] = arr[i + b] + 1;
arr[i + c + 1] = arr[i + c + 1] -1;
for i in range(1, x-1):
arr[i] += arr[i - 1];
print(arr[i], end = " ");
if __name__ == '__main__':
a = 1;
b = 2;
c = 2;
CountWays(a, b, c);
|
C#
using System;
class GFG{
static void CountWays(int a, int b, int c)
{
int x = b + c + 2;
int []arr = new int[x];
for(int i = 1; i <= b; i++)
{
arr[i + b]++;
arr[i + c + 1]--;
}
for(int i = 1; i < x - 1; i++)
{
arr[i] += arr[i - 1];
Console.Write(arr[i] + " ");
}
Console.WriteLine();
}
public static void Main()
{
int a = 1;
int b = 2;
int c = 2;
CountWays(a, b, c);
}
}
|
Javascript
<script>
function CountWays(a, b, c)
{
let x = b + c + 2;
let arr = new Array(x);
arr.fill(0);
for(let i = 1; i <= b; i++)
{
arr[i + b]++;
arr[i + c + 1]--;
}
for(let i = 1; i < x - 1; i++)
{
arr[i] += arr[i - 1];
document.write(arr[i] + " ");
}
document.write("</br>");
}
let a = 1;
let b = 2;
let c = 2;
CountWays(a, b, c);
</script>
|
Time complexity: O(b+c), as we are using loop to traverse b+c times.
Auxiliary Space: O(b+c), as we are using extra space.