Build Array of length N such that exactly X Subarrays have positive sums
Last Updated :
24 Jul, 2023
Given integers N and X, the task is to construct an array of size N such that exactly X subarrays have positive sums and other subarrays have negative sums.
Note: If multiple subarrays are possible, print any of them.
Examples:
Input: N = 3, X = 2
Output: 2 -1 -2
Explanation: [0, 0] and [0, 1] subarrays have positive sums while other subarrays have negative sums.
Input: N = 5, X = 8
Output: 10 3 -1 -1 -2
Explanation: Any subarray starting with index 1 has positive sum and subarray [2, 2], [2, 3], [2, 4] has also positive sum while others subarrays has negative sums
Approach: To solve the problem follow the below idea:
We will iterate from i = 0 to n-1 and will do the following to build the array (say a[]):
- If X ≥ N – i Then we will assign 2N to a[i] and reduce X = X – (N – i) because we will assign a[i+1] to a[n-1] ≥ -2, so we can make positive sum (N – i) subarrays and
- If X ≤ N – i then we will assign X to a[i] and we will assign -1 to index [i+1, i+(X-1)]. This will make exactly X positive sum subarrays because we will assign -2 to in the range [i+X, N-1].
Illustration:
Follow the below illustration for a better understanding:
Consider N = 5, X = 8
- We will start iterating from index 0 to n-1( we are using 0-based indexing here).
- At index 0, X ≥ N-i, so we will assign 2N to index i. and X becomes 8 – (5 – 0) = 3
- At index 1, X < N-i, so we will assign X to a[i].
- Now from the index (1+1 = 2) to (1+3-1) = 3 will be assigned with -1.
- So at index 2, a[i] = -2.
- At index 3 also assign -1 to a[i].
- At index 4, it is out of range [2, 3], so all indices from hereafter will have the value -2. Assign -2 to a[i].
Below are the steps to implement the approach:
- First start iterating from index 0 to n-1 ( 0-based indexing ).
- Assign flag = false if X > 0 and assign flag = true if X = 0.
- At any index, X ≥ N-i, so we will assign 2N to a[i] and reduce x to x-(n-i) and if X = 0, then assign flag = true.
- At any index, X < N-i, so we will assign X to a[i] and assign flag=true.
- If at any index, flag = true and x > 1, then we will assign -1 to a[i] and reduce x to x-1.
- If at any index, flag = true and X = 1, then we will assign -2 to a[i].
- After iterating, print the final array.
Below is the code for the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void xsubarrays(int n, int x)
{
bool flag = false;
if (x == 0) {
flag = true;
}
int arr[n];
for (int i = 0; i < n; i++) {
if (flag) {
if (x > 1) {
arr[i] = -1;
x -= 1;
}
else {
arr[i] = -2;
}
}
else if (x >= n - i) {
arr[i] = n * 2;
x -= (n - i);
if (x == 0) {
flag = true;
}
}
else {
arr[i] = x;
flag = true;
}
}
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
cout << endl;
}
int main()
{
int n = 5, x = 8;
xsubarrays(n, x);
return 0;
}
|
Java
import java.util.*;
public class Main {
static void xsubarrays(int n, int x)
{
boolean flag = false;
if (x == 0) {
flag = true;
}
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
if (flag) {
if (x > 1) {
arr[i] = -1;
x -= 1;
}
else {
arr[i] = -2;
}
}
else if (x >= n - i) {
arr[i] = n * 2;
x -= (n - i);
if (x == 0) {
flag = true;
}
}
else {
arr[i] = x;
flag = true;
}
}
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}
public static void main(String[] args)
{
int n = 5, x = 8;
xsubarrays(n, x);
}
}
|
Python3
def xsubarrays(n, x):
flag = False
if x == 0:
flag = True
arr = [0] * n
for i in range(n):
if flag:
if x > 1:
arr[i] = -1
x -= 1
else:
arr[i] = -2
elif x >= n - i:
arr[i] = n * 2
x -= (n - i)
if x == 0:
flag = True
else:
arr[i] = x
flag = True
print(*arr)
n = 5
x = 8
xsubarrays(n, x)
|
C#
using System;
class GFG
{
static void XSubarrays(int n, int x)
{
bool flag = false;
if (x == 0)
{
flag = true;
}
int[] arr = new int[n];
for (int i = 0; i < n; i++)
{
if (flag)
{
if (x > 1)
{
arr[i] = -1;
x -= 1;
}
else
{
arr[i] = -2;
}
}
else if (x >= n - i)
{
arr[i] = n * 2;
x -= (n - i);
if (x == 0)
{
flag = true;
}
}
else
{
arr[i] = x;
flag = true;
}
}
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
Console.WriteLine();
}
static void Main()
{
int n = 5, x = 8;
XSubarrays(n, x);
}
}
|
Javascript
function xsubarrays(n, x) {
let flag = false;
if (x === 0) {
flag = true;
}
let arr = [];
for (let i = 0; i < n; i++) {
if (flag) {
if (x > 1) {
arr[i] = -1;
x -= 1;
}
else {
arr[i] = -2;
}
}
else if (x >= n - i) {
arr[i] = n * 2;
x -= (n - i);
if (x === 0) {
flag = true;
}
}
else {
arr[i] = x;
flag = true;
}
}
for (let i = 0; i < n; i++) {
console.log(arr[i] + " ");
}
console.log();
}
let n = 5, x = 8;
xsubarrays(n, x);
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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