Longest Subarray with Equal "1" and "0"

Problem: Given an array that only contains "1" and "0", find the longest subarray which contains equal number of "1" and "0".

Solution: With hash table, we can have a O(N) solution. The detail is as follow:

• First convert all "0" to "-1", then calculate c[i] = sum(a[0], ... , a[i]). It takes O(N) to calculate all the c[i].
• Then our task is to find a c[i] and a c[j] such that  c[i] = c[j]  and |j-i| is maximum. With a hash table, we can finish this job by doing a linear scan with a time complexity of   O(N).  // 用hashtable存储具有相同key的position值
• There is a special case you need to handle. When c[N-1] = 0 (assume N is the size of a), the longest subarray is just a itself.

Idea: For every sum value record its leftmost occrrence in the left array and rightmost occrnce in the right array.such that in sum value at ith index == sum value at jth index it means that the subarray from index i to j has = no. of 1's and 0's.

Code:

#include<stdio.h>#include<conio.h>int main(){    int arr[100], rite[200],lft[200],len[200],sum[100], n,i,tmp, lnth=0;    scanf("%d",&n);    printf("%d\n",n);    scanf("%d",&tmp);    if(tmp==0)    sum[0] = -1;    else    sum[0] = tmp;    for(i=1;i<n;i++)    {                     scanf("%d",&tmp);                     arr[i] = tmp;                     if(tmp==0)                     sum[i] = sum[i-1] - 1;                      else                     sum[i] = sum[i-1] + 1;                     if(sum[i]==0)                     lnth = i;                    // printf("%d\n",sum[i]);                     }   //printf("sum = %d\n",sum[n-1]);                     for(i=0;i<(2*n-1);i++)   {                         lft[i] = 0;                         rite[i] = 0;                         }                     for(i=0;i<n;i++)   {                   if(sum[i]>0)                       rite[sum[i]+n-1] = i;                   else if(sum[i]<0)                   rite[sum[i]+n] = i;                   }   for(i=n-1;i>=0;i--)   {                   if(sum[i]>0)                       lft[sum[i]+n-1] = i;                   else if(sum[i]<0)                   lft[sum[i]+n] = i;                      }   for(i=0;i<(2*n-1);i++)   {                         if(lnth<(rite[i]-lft[i]))                         lnth = (rite[i]-lft[i]);                         }    printf("length = %d",lnth);    getch();    return 0;}

Ref:

http://tristan-interview.blogspot.com/2012/01/longest-subarray-with-equal-1-and-0.html

http://www.thealgorithmist.com/showthread.php/146-Array-of-0-s-and-1-s-Find-max-subarray-of-0-s-1-s