Longest Span with same Sum in two Binary arrays - GeeksforGeeks
Given two binary arrays arr1[] and arr2[] of same size n. Find length of the longest common span (i, j) where j >= i such that arr1[i] + arr1[i+1] + …. + arr1[j] = arr2[i] + arr2[i+1] + …. + arr2[j].
Time Complexity: Θ(n)
Auxiliary Space: Θ(n)
The idea is based on below observations.
Method 1 (Simple Solution)
One by one by consider same subarrays of both arrays. For all subarrays, compute sums and if sums are same and current length is more than max length, then update max length.
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Given two binary arrays arr1[] and arr2[] of same size n. Find length of the longest common span (i, j) where j >= i such that arr1[i] + arr1[i+1] + …. + arr1[j] = arr2[i] + arr2[i+1] + …. + arr2[j].
Input: arr1[] = {0, 1, 0, 0, 0, 0}; arr2[] = {1, 0, 1, 0, 0, 1}; Output: 4 The longest span with same sun is from index 1 to 4.Method 2 (Using Auxiliary Array)
Time Complexity: Θ(n)
Auxiliary Space: Θ(n)
The idea is based on below observations.
- Since there are total n elements, maximum sum is n for both arrays.
- Difference between two sums varies from -n to n. So there are total 2n + 1 possible values of difference.
- If differences between prefix sums of two arrays become same at two points, then subarrays between these two points have same sum.
Below is Complete Algorithm.
- Create an auxiliary array of size 2n+1 to store starting points of all possible values of differences (Note that possible values of differences vary from -n to n, i.e., there are total 2n+1 possible values)
- Initialize starting points of all differences as -1.
- Initialize maxLen as 0 and prefix sums of both arrays as 0, preSum1 = 0, preSum2 = 0
- Travers both arrays from i = 0 to n-1.
- Update prefix sums: preSum1 += arr1[i], preSum2 += arr2[i]
- Compute difference of current prefix sums: curr_diff = preSum1 – preSum2
- Find index in diff array: diffIndex = n + curr_diff // curr_diff can be negative and can go till -n
- If curr_diff is 0, then i+1 is maxLen so far
- Else If curr_diff is seen first time, i.e., starting point of current diff is -1, then update starting point as i
- Else (curr_diff is NOT seen first time), then consider i as ending point and find length of current same sum span. If this length is more, then update maxLen
- Return maxLen
int
longestCommonSum(
bool
arr1[],
bool
arr2[],
int
n)
{
// Initialize result
int
maxLen = 0;
// Initialize prefix sums of two arrays
int
preSum1 = 0, preSum2 = 0;
// Create an array to store staring and ending
// indexes of all possible diff values. diff[i]
// would store starting and ending points for
// difference "i-n"
int
diff[2*n+1];
// Initialize all starting and ending values as -1.
memset
(diff, -1,
sizeof
(diff));
// Traverse both arrays
for
(
int
i=0; i<n; i++)
{
// Update prefix sums
preSum1 += arr1[i];
preSum2 += arr2[i];
// Comput current diff and index to be used
// in diff array. Note that diff can be negative
// and can have minimum value as -1.
int
curr_diff = preSum1 - preSum2;
int
diffIndex = n + curr_diff;
// If current diff is 0, then there are same number
// of 1's so far in both arrays, i.e., (i+1) is
// maximum length.
if
(curr_diff == 0)
maxLen = i+1;
// If current diff is seen first time, then update
// starting index of diff.
else
if
( diff[diffIndex] == -1)
diff[diffIndex] = i;
// Current diff is already seen
else
{
// Find lenght of this same sum common span
int
len = i - diff[diffIndex];
// Update max len if needed
if
(len > maxLen)
maxLen = len;
}
}
return
maxLen;
}
Method 1 (Simple Solution)
One by one by consider same subarrays of both arrays. For all subarrays, compute sums and if sums are same and current length is more than max length, then update max length.
int
longestCommonSum(
bool
arr1[],
bool
arr2[],
int
n)
{
// Initialize result
int
maxLen = 0;
// One by one pick all possible starting points
// of subarrays
for
(
int
i=0; i<n; i++)
{
// Initialize sums of current subarrays
int
sum1 = 0, sum2 = 0;
// Conider all points for starting with arr[i]
for
(
int
j=i; j<n; j++)
{
// Update sums
sum1 += arr1[j];
sum2 += arr2[j];
// If sums are same and current length is
// more than maxLen, update maxLen
if
(sum1 == sum2)
{
int
len = j-i+1;
if
(len > maxLen)
maxLen = len;
}
}
}
return
maxLen;
}
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