Given three integers N, S, and K, the task is to create an array of N positive integers such that the bitwise OR of any two consecutive elements from the array is odd and there are exactly K subarrays with a sum equal to S where 1 ? K ? N / 2.
Examples:
Input: N = 4, K = 2, S = 6
Output: 6 7 6 7
Here, there are exactly 2 subarray {6} and {6}
whose sum is 6 and the bitwise OR of
any adjacent elements is odd.
Input: N = 8, K = 3, S = 12
Output: 12 13 12 13 12 13 13 13
Approach:
- Observe a pattern here {S, P, S, P, S, P, ..., P, P, P, P}.
- Here P is an odd number > S and after every S there is an occurrence of P. It is known that the bitwise OR with an odd number is always an odd number, so it is confirmed that bitwise OR of each adjacent element is an odd number.
- Now, put exactly K number of S in the above pattern of the array.
- Except for S all the elements (which are P) are greater than S, so there can not be any subarray whose sum is exactly S other than those K subarrays.
Below is the implementation of the above approach:
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to print the
// contents of an array
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to generate and print
// the required array
void findArray(int n, int k, int s)
{
// Initially all the positions are empty
int vis[n] = { 0 };
// To store the count of positions
// i such that arr[i] = s
int cnt = 0;
// To store the final array elements
int arr[n];
for (int i = 0; i < n && cnt < k; i += 2) {
// Set arr[i] = s and the gap between
// them is exactly 2 so in for loop
// we use i += 2
arr[i] = s;
// Mark the i'th position as visited
// as we put arr[i] = s
vis[i] = 1;
// Increment the count
cnt++;
}
int val = s;
// Finding the next odd number after s
if (s % 2 == 0)
val++;
else
val = val + 2;
for (int i = 0; i < n; i++) {
if (vis[i] == 0) {
// If the i'th position is not visited
// it means we did not put any value
// at position i so we put 1 now
arr[i] = val;
}
}
// Print the final array
printArr(arr, n);
}
// Driver code
int main()
{
int n = 8, k = 3, s = 12;
findArray(n, k, s);
return 0;
}
// Java implementation of the approach
class GFG
{
// Utility function to print the
// contents of an array
static void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Function to generate and print
// the required array
static void findArray(int n, int k, int s)
{
// Initially all the positions are empty
int vis[] = new int[n] ;
// To store the count of positions
// i such that arr[i] = s
int cnt = 0;
// To store the final array elements
int arr[] = new int[n];
for (int i = 0; i < n && cnt < k; i += 2)
{
// Set arr[i] = s and the gap between
// them is exactly 2 so in for loop
// we use i += 2
arr[i] = s;
// Mark the i'th position as visited
// as we put arr[i] = s
vis[i] = 1;
// Increment the count
cnt++;
}
int val = s;
// Finding the next odd number after s
if (s % 2 == 0)
val++;
else
val = val + 2;
for (int i = 0; i < n; i++)
{
if (vis[i] == 0)
{
// If the i'th position is not visited
// it means we did not put any value
// at position i so we put 1 now
arr[i] = val;
}
}
// Print the final array
printArr(arr, n);
}
// Driver code
public static void main (String[] args)
{
int n = 8, k = 3, s = 12;
findArray(n, k, s);
}
}
// This code is contributed by AnkitRai01
# Python3 implementation of the approach
# Utility function to print the
# contents of an array
def printArr(arr, n) :
for i in range(n) :
print(arr[i], end= " ");
# Function to generate and print
# the required array
def findArray(n, k, s) :
# Initially all the positions are empty
vis = [0] * n;
# To store the count of positions
# i such that arr[i] = s
cnt = 0;
# To store the final array elements
arr = [0] * n;
i = 0;
while (i < n and cnt < k) :
# Set arr[i] = s and the gap between
# them is exactly 2 so in for loop
# we use i += 2
arr[i] = s;
# Mark the i'th position as visited
# as we put arr[i] = s
vis[i] = 1;
# Increment the count
cnt += 1;
i += 2;
val = s;
# Finding the next odd number after s
if (s % 2 == 0) :
val += 1;
else :
val = val + 2;
for i in range(n) :
if (vis[i] == 0) :
# If the i'th position is not visited
# it means we did not put any value
# at position i so we put 1 now
arr[i] = val;
# Print the final array
printArr(arr, n);
# Driver code
if __name__ == "__main__" :
n = 8; k = 3; s = 12;
findArray(n, k, s);
# This code is contributed by AnkitRai01
// C# implementation of the approach
using System;
class GFG
{
// Utility function to print the
// contents of an array
static void printArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Function to generate and print
// the required array
static void findArray(int n, int k, int s)
{
// Initially all the positions are empty
int []vis = new int[n] ;
// To store the count of positions
// i such that arr[i] = s
int cnt = 0;
// To store the final array elements
int []arr = new int[n];
for (int i = 0; i < n && cnt < k; i += 2)
{
// Set arr[i] = s and the gap between
// them is exactly 2 so in for loop
// we use i += 2
arr[i] = s;
// Mark the i'th position as visited
// as we put arr[i] = s
vis[i] = 1;
// Increment the count
cnt++;
}
int val = s;
// Finding the next odd number after s
if (s % 2 == 0)
val++;
else
val = val + 2;
for (int i = 0; i < n; i++)
{
if (vis[i] == 0)
{
// If the i'th position is not visited
// it means we did not put any value
// at position i so we put 1 now
arr[i] = val;
}
}
// Print the final array
printArr(arr, n);
}
// Driver code
public static void Main()
{
int n = 8, k = 3, s = 12;
findArray(n, k, s);
}
}
// This code is contributed by AnkitRai01
<script>
// javascript implementation of the approach
// Utility function to print the
// contents of an array
function printArr(arr , n) {
for (i = 0; i < n; i++)
document.write(arr[i] + " ");
}
// Function to generate and print
// the required array
function findArray(n , k , s) {
// Initially all the positions are empty
var vis = Array(n).fill(0);
// To store the count of positions
// i such that arr[i] = s
var cnt = 0;
// To store the final array elements
var arr = Array(n).fill(0);
for (i = 0; i < n && cnt < k; i += 2) {
// Set arr[i] = s and the gap between
// them is exactly 2 so in for loop
// we use i += 2
arr[i] = s;
// Mark the i'th position as visited
// as we put arr[i] = s
vis[i] = 1;
// Increment the count
cnt++;
}
var val = s;
// Finding the next odd number after s
if (s % 2 == 0)
val++;
else
val = val + 2;
for (i = 0; i < n; i++) {
if (vis[i] == 0) {
// If the i'th position is not visited
// it means we did not put any value
// at position i so we put 1 now
arr[i] = val;
}
}
// Print the final array
printArr(arr, n);
}
// Driver code
var n = 8, k = 3, s = 12;
findArray(n, k, s);
// This code contributed by umadevi9616
</script>
Output:
12 13 12 13 12 13 13 13
Time Complexity: O(N)
Auxiliary Space: O(N) it is using extra space for arrays