Given an array of 0's and 1's. only, find the maximum length of the
subarray such that the number of 0's and 1's in that subarray are equal.
Examples
Input: arr[] = {1, 0, 1, 1, 1, 0, 0}
Output: 1 to 6 (Starting and Ending indexes of output subarray)
Input: arr[] = {1, 1, 1, 1}
Output: No such subarray
Input: arr[] = {0, 0, 1, 1, 0}
Output: 0 to 3 Or 1 to 4
Solution:
We can change 0 to -1 on the original array.
Define a sum array S[i] = a[1] + ... + a[i] (a being the input)
Now We want to find indices i and j such that a[i]+...+a[j] = 0
and that is the same as
S[j]-S[i-1] = 0
S[j] = S[i-1]
now We can hash these values (they range from -n to n, so we can do this in O(n) memory)and find the indices in O(n). We can also take an array of size
2n+1 and store the first index containing the value seen and compute
difference whenever u see the value again ..
Following is a solution that uses O(n) extra space and solves the problem in O(n) time complexity.
Let input array be arr[] of size n and maxsize be the size of output subarray.
1) Consider all 0 values as -1. The problem now reduces to find out the maximum length subarray with sum = 0.
2) Create a temporary array sumleft[] of size n. Store the sum of all elements from arr[0] to arr[i] in sumleft[i]. This can be done in O(n) time.
3) There are two cases, the output subarray may start from 0th index or may start from some other index. We will return the max of the values obtained by two cases.
4) To find the maximum length subarray starting from 0th index, scan the sumleft[] and find the maximum i where sumleft[i] = 0.
5) Now, we need to find the subarray where subarray sum is 0 and start index is not 0. This problem is equivalent to finding two indexes i & j in sumleft[] such that sumleft[i] = sumleft[j] and j-i is maximum. To solve this, we can create a hash table with size = max-min+1 where min is the minimum value in the sumleft[] and max is the maximum value in the sumleft[]. The idea is to hash the leftmost occurrences of all different values in sumleft[]. The size of hash is chosen as max-min+1 because there can be these many different possible values in sumleft[]. Initialize all values in hash as -1
6) To fill and use hash[], traverse sumleft[] from 0 to n-1. If a value is not present in hash[], then store its index in hash. If the value is present, then calculate the difference of current index of sumleft[] and previously stored value in hash[]. If this difference is more than maxsize, then update the maxsize.
7) To handle corner cases (all 1s and all 0s), we initialize maxsize as -1. If the maxsize remains -1, then print there is no such subarray.
#include <stdio.h>
#include <stdlib.h>
int max(int a, int b) { return a>b? a: b; }
int findSubArray(int arr[], int n)
{
int maxsize = -1, startindex;
int sumleft[n];
int min, max; // For min and max values in sumleft[]
int i;
// Fill sumleft array and get min and max values in it.
// Consider 0 values in arr[] as -1
sumleft[0] = ((arr[0] == 0)? -1: 1);
min = arr[0]; max = arr[0];
for (i=1; i<n; i++)
{
sumleft[i] = sumleft[i-1] + ((arr[i] == 0)? -1: 1);
if (sumleft[i] < min)
min = sumleft[i];
if (sumleft[i] > max)
max = sumleft[i];
}
// Now calculate the max value of j - i such that sumleft[i] = sumleft[j].
// The idea is to create a hash table to store indexes of all visited values.
// If you see a value again, that it is a case of sumleft[i] = sumleft[j]. Check
// if this j-i is more than maxsize.
// The optimum size of hash will be max-min+1 as these many different values
// of sumleft[i] are possible. Since we use optimum size, we need to shift
// all values in sumleft[] by min before using them as an index in hash[].
int hash[max-min+1];
// Initialize hash table
for (i=0; i<max-min+1; i++)
hash[i] = -1;
for (i=0; i<n; i++)
{
// Case 1: when the subarray starts from index 0
if (sumleft[i] == 0)
{
maxsize = i+1;
startindex = 0;
}
// Case 2: fill hash table value. If already filled, then use it
if (hash[sumleft[i]-min] == -1)
hash[sumleft[i]-min] = i;
else
{
if ( (i - hash[sumleft[i]-min]) > maxsize )
{
maxsize = i - hash[sumleft[i]-min];
startindex = hash[sumleft[i]-min] + 1;
}
}
}
if ( maxsize == -1 )
printf("No such subarray");
else
printf("%d to %d", startindex, startindex+maxsize-1);
return maxsize;
}
int main()
{
int arr[] = {1, 0, 0, 1, 0, 1, 1};
int size = sizeof(arr)/sizeof(arr[0]);
findSubArray(arr, size);
return 0;
}
Time Complexity: O(n)
Auxiliary Space: O(n)
http://www.geeksforgeeks.org/largest-subarray-with-equal-number-of-0s-and-1s/
