# Share my super easy solution O(n) time and O(1) space

• The idea is pretty simple think about 1,2,3,4,5...,

if we stop at 3 we have 1 option;
if we stop at 4 we have 3 options;
if we stop at 5 we have 5 option;
.
.
.

so if we know how many options we have now and how many options we have before: we can have this simple formula:

next = current*2 + 1 - previous;

The reason is pretty straightforward, thinking about from 3->4 -> 5:
At 3 we have 1,2,3
At 4 we have 1,2,3 | 2,3,4 | 1,2,3,4
At 5 we have 1,2,3 | 2,3,4 | 3,4,5 | 1,2,3,4 | 2,3,4,5 | 1,2,3,4,5

if we want to calculate how many options we have at 5, we know 1,2,3,4 have 3 options and 2,3,4,5 also have 3 options and 1,2,3,4,5, so now we have 7 options but we have a overlap 2,3,4 so we need to minus the overlap, which equals how many options we have at 3.

``````public class Solution {
public int numberOfArithmeticSlices(int[] A) {
int len = A.length;
if(len < 3) return 0;

int diff = A[1] - A[0];
int curDiff;

int result = 0;
int current = 0;
int prev = 0;
for(int i = 2 ; i < len ; i++){
curDiff = A[i] - A[i-1];
if(curDiff == diff){
if(current == 0) current = 1;
else{
int temp = current;
current = current*2+1-prev;
prev = temp;
}
}else{
diff = curDiff;
result += current;
current = 0;
}
}

result += current;
return result;
}
}
``````

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