Given an integer n, return any array containing n unique integers such that they add up to 0.
Example 1:
Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].
Approach:- So there will be array of either even or odd length
if array is of length 4, the array will be[-1, -2, 1, 2]
if array is of length 5, the array will be[0, -1, -2, 1, 2]
okay, seems pretty straightforward from here on
public int[] sumZero(int n) {
int []ans = new int[n];
int currentAns = 1;
int currentPointer = 0;
if(n%2==1){
ans[currentPointer++] = 0;
}
while(currentPointer<n){
ans[currentPointer++] = currentAns;
ans[currentPointer++] = - currentAns;
currentAns++;
}
return ans;
}
Time comlexity - O(n) Space comlexity - O(n)