Given an array arr[] of size N, the task is to count the number of arrays having at least K elements greater than the XOR of all array elements, generated by performing the following operations X times.
- Select either first or last element from the given array.
- Either increment the selected element by 1 or delete the selected element.
Examples:
Input: arr[] = {10, 2, 10, 5}, X = 3, K = 3
Output: 1
Explanation:
XOR of the given array = 7. The only possible array satisfying the condition is {10, 2, 10, 8}, obtained by incrementing the last array element thrice.Input: arr[] = {3, 3, 4}, X = 3, K = 2
Output: 3
Approach: The idea is to use the backtracking approach to recursively try out all the possible moves and increment the count when the required array is obtained. Possible moves are:
- Initialize a variable, say xorValue, to calculate the XOR of the original array.
- Initialize a variable, say count, to store the final count of the required arrays.
- Recursively try all the following four possibilities and increment the count when the required array is obtained:
- Delete the first element of the array.
- Delete the last element of the array.
- Increment the first array element by one.
- Increment last array element by one.
- Print the final count as the answer.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Stores the final answer
int ans = 0;
// Utility function to count arrays
// having at least K elements exceeding
// XOR of all given array elements
void countArraysUtil(vector<int>& arr,
int X, int K,
int xorVal)
{
// If no operations are left
if (X == 0) {
// Stores the count of
// possible arrays
int cnt = 0;
// Count array elements are
// greater than XOR
for (int i = 0; i < arr.size(); i++) {
if (arr[i] > xorVal)
cnt++;
}
if (cnt >= K)
ans++;
return;
}
// Stores first element
int temp = arr[0];
// Delete first element
arr.erase(arr.begin());
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Insert first element into vector
arr.insert(arr.begin(), temp);
// Stores the last element
temp = arr.back();
// Remove last element from vector
arr.pop_back();
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Push last element into vector
arr.push_back(temp);
// Increment first element
arr[0]++;
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Decrement first element
arr[0]--;
// Increment last element
arr[arr.size() - 1]++;
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Decrement last element
arr[arr.size() - 1]--;
}
// Function to find the count of
// arrays having atleast K elements
// greater than XOR of array
void countArrays(vector<int>& arr,
int X, int K)
{
// Stores the XOR value
// of original array
int xorVal = 0;
// Traverse the vector
for (int i = 0; i < arr.size(); i++)
xorVal = xorVal ^ arr[i];
countArraysUtil(arr, X, K, xorVal);
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
// Given vector
vector<int> arr = { 10, 2, 10, 5 };
// Given value of X & K
int X = 3, K = 3;
countArrays(arr, X, K);
return 0;
}
// Java program for the above approach
import java.util.ArrayList;
class GFG{
// Stores the final answer
static int ans = 0;
// Utility function to count arrays
// having at least K elements exceeding
// XOR of all given array elements
public static void countArraysUtil(ArrayList<Integer> arr,
int X, int K,
int xorVal)
{
// If no operations are left
if (X == 0) {
// Stores the count of
// possible arrays
int cnt = 0;
// Count array elements are
// greater than XOR
for (int i = 0; i < arr.size(); i++) {
if (arr.get(i) > xorVal)
cnt++;
}
if (cnt >= K)
ans++;
return;
}
// Stores first element
int temp = arr.get(0);
// Delete first element
arr.remove(0);
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Insert first element into vector
arr.add(0, temp);
// Stores the last element
temp = arr.get(arr.size() - 1);
// Remove last element from vector
arr.remove(arr.size() - 1);
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Push last element into vector
arr.add(temp);
// Increment first element
arr.set(0, arr.get(0) + 1);
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Decrement first element
arr.set(0, arr.get(0) - 1);
// Increment last element
arr.set(arr.size() - 1, arr.get(arr.size() - 1) + 1);
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Decrement last element
arr.set(arr.size() - 1, arr.get(arr.size() - 1) - 1);
}
// Function to find the count of
// arrays having atleast K elements
// greater than XOR of array
public static void countArrays(ArrayList<Integer> arr,
int X, int K)
{
// Stores the XOR value
// of original array
int xorVal = 0;
// Traverse the vector
for (int i = 0; i < arr.size(); i++)
xorVal = xorVal ^ arr.get(i);
countArraysUtil(arr, X, K, xorVal);
// Print the answer
System.out.println(ans);
}
// Driver Code
public static void main(String arg[])
{
// Given vector
int[] input = {10, 2, 10, 5};
// Convert the input as ArrayList
ArrayList<Integer> arr = new ArrayList<Integer>();
for(int i : input){
arr.add(i);
}
// Given value of X & K
int X = 3, K = 3;
countArrays(arr, X, K);
}
}
// This code is contributed by gfgking.
# Python program for the above approach
# Stores the final answer
ans = 0
# Utility function to count arrays
# having at least K elements exceeding
# XOR of all given array elements
def countArraysUtil( arr, X, K, xorVal):
global ans
# If no operations are left
if (X == 0):
# Stores the count of
# possible arrays
cnt = 0
# Count array elements are
# greater than XOR
for i in range(len(arr)):
if (arr[i] > xorVal):
cnt += 1
if (cnt >= K):
ans += 1
return
# Stores first element
temp = arr[0]
# Delete first element
arr.pop(0)
# Recursive call
countArraysUtil(arr, X - 1, K, xorVal)
# Insert first element into vector
arr.insert(0, temp)
# Stores the last element
temp = arr[-1]
# Remove last element from vector
arr.pop()
# Recursive call
countArraysUtil(arr, X - 1, K, xorVal)
# Push last element into vector
arr.append(temp)
# Increment first element
arr[0] += 1
# Recursive call
countArraysUtil(arr, X - 1,K, xorVal)
# Decrement first element
arr[0] -= 1
# Increment last element
arr[len(arr) - 1] += 1
# Recursive call
countArraysUtil(arr, X - 1, K, xorVal)
# Decrement last element
arr[len(arr) - 1] -= 1
# Function to find the count of
# arrays having atleast K elements
# greater than XOR of array
def countArrays(arr, X, K):
# Stores the XOR value
# of original array
xorVal = 0
# Traverse the vector
for i in range(len(arr)):
xorVal = xorVal ^ arr[i]
countArraysUtil(arr, X, K, xorVal)
# Print the answer
print(ans)
# Driver Code
# Given vector
arr = [ 10, 2, 10, 5 ]
# Given value of X & K
X = 3
K = 3
countArrays(arr, X, K)
# This code is contributed by rohitsingh07052.
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Stores the final answer
static int ans = 0;
// Utility function to count arrays
// having at least K elements exceeding
// XOR of all given array elements
static void countArraysUtil(List<int> arr, int X, int K,
int xorVal)
{
// If no operations are left
if (X == 0) {
// Stores the count of
// possible arrays
int cnt = 0;
// Count array elements are
// greater than XOR
for (int i = 0; i < arr.Count; i++) {
if (arr[i] > xorVal)
cnt++;
}
if (cnt >= K)
ans++;
return;
}
// Stores first element
int temp = arr[0];
// Delete first element
arr.RemoveAt(0);
// Recursive call
countArraysUtil(arr, X - 1, K, xorVal);
// Insert first element into vector
arr.Insert(0, temp);
// Stores the last element
temp = arr[arr.Count - 1];
// Remove last element from vector
arr.RemoveAt(arr.Count - 1);
// Recursive call
countArraysUtil(arr, X - 1, K, xorVal);
// Push last element into vector
arr.Add(temp);
// Increment first element
arr[0]++;
// Recursive call
countArraysUtil(arr, X - 1, K, xorVal);
// Decrement first element
arr[0]--;
// Increment last element
arr[arr.Count - 1]++;
// Recursive call
countArraysUtil(arr, X - 1, K, xorVal);
// Decrement last element
arr[arr.Count - 1]--;
}
// Function to find the count of
// arrays having atleast K elements
// greater than XOR of array
static void countArrays(List<int> arr, int X, int K)
{
// Stores the XOR value
// of original array
int xorVal = 0;
// Traverse the vector
for (int i = 0; i < arr.Count; i++)
xorVal = xorVal ^ arr[i];
countArraysUtil(arr, X, K, xorVal);
// Print the answer
Console.Write(ans);
}
// Driver Code
public static void Main()
{
// Given vector
List<int> arr = new List<int>() { 10, 2, 10, 5 };
// Given value of X & K
int X = 3, K = 3;
countArrays(arr, X, K);
}
}
// This code is contributed by chitranayal.
<script>
// JavaScript program for the above approach
// Stores the final answer
let ans = 0;
// Utility function to count arrays
// having at least K elements exceeding
// XOR of all given array elements
function countArraysUtil(arr,X,K,xorVal)
{
// If no operations are left
if (X == 0) {
// Stores the count of
// possible arrays
let cnt = 0;
// Count array elements are
// greater than XOR
for (let i = 0; i < arr.length; i++) {
if (arr[i] > xorVal)
cnt++;
}
if (cnt >= K)
ans++;
return;
}
// Stores first element
let temp = arr[0];
// Delete first element
arr.shift();
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Insert first element into vector
arr.unshift(temp);
// Stores the last element
temp = arr[arr.length-1];
// Remove last element from vector
arr.pop();
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Push last element into vector
arr.push(temp);
// Increment first element
arr[0]++;
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Decrement first element
arr[0]--;
// Increment last element
arr[arr.length - 1]++;
// Recursive call
countArraysUtil(arr, X - 1,
K, xorVal);
// Decrement last element
arr[arr.length - 1]--;
}
// Function to find the count of
// arrays having atleast K elements
// greater than XOR of array
function countArrays(arr,X,K)
{
// Stores the XOR value
// of original array
let xorVal = 0;
// Traverse the vector
for (let i = 0; i < arr.length; i++)
xorVal = xorVal ^ arr[i];
countArraysUtil(arr, X, K, xorVal);
// Print the answer
document.write(ans);
}
// Driver Code
let arr=[10, 2, 10, 5];
// Given value of X & K
let X = 3, K = 3;
countArrays(arr, X, K);
// This code is contributed by avanitrachhadiya2155
</script>
Output:
1
Time Complexity: O(4K * N)
Auxiliary Space: O(1)