# Explicit Java solution (O(n) time and O(1) space)

• We can solve this by just keeping track of what the current difference is and how many consecutive elements with this difference we have already seen.

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

int counter = 0;    // counts number of all arithmetic slices
// asume we have solution for 1...i-1.
// Then we can check diff = A[i]-A[i-1] and see how many previous sequences it extends
int runningDiff = A[1] - A[0];
int runningCount = 1;   // how many with the same runningDiff have we encountered _consecutively_
for(int i = 2; i < n; i++) {
int curDiff = A[i] - A[i-1];
if(curDiff == runningDiff) {
runningCount++;
// runningCount of j implies there is an array of length j+1
// increasing the previous array by one element increases the number of all 3,4,5,...,j-tuples
// by one and introduces a new j+1 tuple.
// So, in total we found 3,...,(j+1) = j-1 new tuples
if(runningCount >= 2) counter += (runningCount - 1);
} else {
runningDiff = curDiff;
runningCount = 1;
}
}
return counter;
}
}
``````

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