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# Number of ways to obtain each numbers in range [1, b+c] by adding any two numbers in range [a, b] and [b, c]

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++

 `// C++ program to calculate``// the number of ways` `#include ``using` `namespace` `std;` `void` `CountWays(``int` `a, ``int` `b, ``int` `c)``{``    ``int` `x = b + c + 1;``    ``int` `arr[x] = { 0 };` `    ``// Initialising the array with zeros.``    ``// You can do using memset too.``    ``for` `(``int` `i = a; i <= b; i++) {``        ``for` `(``int` `j = b; j <= c; j++) {``            ``arr[i + j]++;``        ``}``    ``}``    ``// Printing the array``    ``for` `(``int` `i = 1; i < x; i++) {``        ``cout << arr[i] << ``" "``;``    ``}``    ``cout << endl;``}``// Driver code``int` `main()``{``    ``int` `a = 1;``    ``int` `b = 2;``    ``int` `c = 2;` `    ``CountWays(a, b, c);` `    ``return` `0;``}`

## Java

 `// Java program to calculate``// the number of ways``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];``    ` `    ``// Initialising the array with zeros.``    ``// You can do using memset too.``    ``for``(``int` `i = a; i <= b; i++)``    ``{``       ``for``(``int` `j = b; j <= c; j++)``       ``{``          ``arr[i + j]++;``       ``}``    ``}``    ` `    ``// Printing the array``    ``for``(``int` `i = ``1``; i < x; i++)``    ``{``       ``System.out.print(arr[i] + ``" "``);``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a = ``1``;``    ``int` `b = ``2``;``    ``int` `c = ``2``;``    ` `    ``CountWays(a, b, c);``}``}` `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 program to calculate``# the number of ways``def` `CountWays(a, b, c):``    ` `    ``x ``=` `b ``+` `c ``+` `1``;``    ``arr ``=` `[``0``] ``*` `x;` `    ``# Initialising the array with zeros.``    ``# You can do using memset too.``    ``for` `i ``in` `range``(a, b ``+` `1``):``        ``for` `j ``in` `range``(b, c ``+` `1``):``            ``arr[i ``+` `j] ``+``=` `1``;` `    ``# Printing the array``    ``for` `i ``in` `range``(``1``, x):``        ``print``(arr[i], end ``=` `" "``);``        ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``a ``=` `1``;``    ``b ``=` `2``;``    ``c ``=` `2``;` `    ``CountWays(a, b, c);``    ` `# This code is contributed by Rajput-Ji`

## C#

 `// C# program to calculate``// the number of ways``using` `System;``class` `GFG{``    ` `public` `static` `void` `CountWays(``int` `a, ``int` `b,``                                    ``int` `c)``{``    ``int` `x = b + c + 1;``    ``int``[] arr = ``new` `int``[x];``    ` `    ``// Initialising the array with zeros.``    ``// You can do using memset too.``    ``for``(``int` `i = a; i <= b; i++)``    ``{``        ``for``(``int` `j = b; j <= c; j++)``        ``{``            ``arr[i + j]++;``        ``}``    ``}``    ` `    ``// Printing the array``    ``for``(``int` `i = 1; i < x; i++)``    ``{``        ``Console.Write(arr[i] + ``" "``);``    ``}``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `a = 1;``    ``int` `b = 2;``    ``int` `c = 2;``    ` `    ``CountWays(a, b, c);``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:

`0 0 1 1`

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.

1. 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.
2. 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++

 `// C++ program to calculate``// the number of ways` `#include ``using` `namespace` `std;` `void` `CountWays(``int` `a, ``int` `b, ``int` `c)``{``    ``// 2 is added because sometimes``    ``// we will decrease the``    ``// value out of bounds.``    ``int` `x = b + c + 2;` `    ``// Initialising the array with zeros.``    ``// You can do using memset too.``    ``int` `arr[x] = { 0 };` `    ``for` `(``int` `i = 1; i <= b; i++) {``        ``arr[i + b]++;``        ``arr[i + c + 1]--;``    ``}` `    ``// Printing the array``    ``for` `(``int` `i = 1; i < x - 1; i++) {``        ``arr[i] += arr[i - 1];``        ``cout << arr[i] << ``" "``;``    ``}``    ``cout << endl;``}` `// Driver code``int` `main()``{``    ``int` `a = 1;``    ``int` `b = 2;``    ``int` `c = 2;` `    ``CountWays(a, b, c);` `    ``return` `0;``}`

## Java

 `// Java program to calculate``// the number of ways``import` `java.io.*;` `class` `GFG{` `static` `void` `CountWays(``int` `a, ``int` `b, ``int` `c)``{``    ` `    ``// 2 is added because sometimes``    ``// we will decrease the``    ``// value out of bounds.``    ``int` `x = b + c + ``2``;` `    ``// Initialising the array with zeros.``    ``int` `arr[] = ``new` `int``[x];` `    ``for``(``int` `i = ``1``; i <= b; i++)``    ``{``       ``arr[i + b]++;``       ``arr[i + c + ``1``]--;``    ``}` `    ``// Printing the array``    ``for``(``int` `i = ``1``; i < x - ``1``; i++)``    ``{``       ``arr[i] += arr[i - ``1``];``       ``System.out.print(arr[i] + ``" "``);``    ``}``    ``System.out.println();``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a = ``1``;``    ``int` `b = ``2``;``    ``int` `c = ``2``;` `    ``CountWays(a, b, c);``}``}` `// This code is contributed by Rohit_ranjan`

## Python3

 `# Python3 program to calculate``# the number of ways``def` `CountWays(a, b, c):``     ` `    ``# 2 is added because sometimes``    ``# we will decrease the``    ``# value out of bounds.``    ``x ``=` `b ``+` `c ``+` `2``;`` ` `    ``# Initialising the array with zeros.``    ``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``;``    ` ` ` `    ``# Printing the array``    ``for` `i ``in` `range``(``1``, x``-``1``):``    ` `       ``arr[i] ``+``=` `arr[i ``-` `1``];``       ``print``(arr[i], end ``=` `" "``);` ` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``     ` `    ``a ``=` `1``;``    ``b ``=` `2``;``    ``c ``=` `2``;`` ` `    ``CountWays(a, b, c);``     ` `# This code is contributed by rock_cool`

## C#

 `// C# program to calculate``// the number of ways``using` `System;``class` `GFG{` `static` `void` `CountWays(``int` `a, ``int` `b, ``int` `c)``{``    ` `    ``// 2 is added because sometimes``    ``// we will decrease the``    ``// value out of bounds.``    ``int` `x = b + c + 2;` `    ``// Initialising the array with zeros.``    ``int` `[]arr = ``new` `int``[x];` `    ``for``(``int` `i = 1; i <= b; i++)``    ``{``        ``arr[i + b]++;``        ``arr[i + c + 1]--;``    ``}` `    ``// Printing the array``    ``for``(``int` `i = 1; i < x - 1; i++)``    ``{``        ``arr[i] += arr[i - 1];``        ``Console.Write(arr[i] + ``" "``);``    ``}``    ``Console.WriteLine();``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `a = 1;``    ``int` `b = 2;``    ``int` `c = 2;` `    ``CountWays(a, b, c);``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``

Output:

`0 0 1 1`

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.