# Length of longest continuous sequence with same sum in given binary arrays

Given two Boolean arrays X and Y, find the length of longest continuous sequence that starts and ends at same index in both arrays and have same sum. In other words, find max(j-i+1) for every j >= i where sum of sub-array X[i, j] is equal to sum of sub-array Y[i, j].

For example, consider below Boolean arrays X[] and Y[]

X: {0, 0, 1, 1, 1, 1}
Y: {0, 1, 1, 0, 1, 0}

The length of longest continuous sequence with same sum is 5 as

X[0, 4]: {0, 0, 1, 1, 1}      (sum = 3)
Y[0, 4]: {0, 1, 1, 0, 1}      (sum = 3)

Naive solution would be to consider every sub-array [i, j] where (j > i) and check if sum of X[i, j] is equal to sum of Y[i, j] or not. If sum is found to be equal and length of the sub-array is more than maximum found so far, we update the result. The time complexity of above solution is O(n2) assuming that sum of each sub-array is computed in constant time.

We can solve this problem in linear time. The idea is to traverse the array and maintain sum of elements of X[] and Y[] till current index and calculate difference between the two sums.

• If the difference is seen for the first time, then we store the difference and current index in a map.

• If difference is seen before and index of previous occurrence is i, then we have found sub-arrays X[i+1, j] and Y[i+1, j] ending at current index j, whose sum of elements is equal. If length of the sub-array is more than maximum found so far, we update the result.

How does this works?

Claim: If difference is seen before and index of previous occurrence is i,
then X[i+1, j] = Y[i+1, j]

We can write previous difference di = X[0, i] – Y[0, i]

Similarly, current difference dj = X[0, j] – Y[0, j] or
dj = (X[0, i] + X[i+1, j]) – (Y[0, i] + Y[i+1, j])

If difference is seen before i.e. (dj = di), then

(X[0, i] + X[i+1, j]) – (Y[0, i] + Y[i+1, j]) = X[0, i] – Y[0, i]
X[i+1, j] – Y[i+1, j] = 0 or
X[i+1, j] == Y[i+1, j]

## C++

Output:

The length of longest continuous sequence with same sum is 5

## Java

Output:

The length of longest continuous sequence with same sum is 5

The time complexity of above solution is O(n) and auxiliary space used by the program is O(n).     (1 votes, average: 5.00 out of 5) Loading...

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