Given an array containing only 0s and 1s, find the largest subarray which contain equal no of 0s and 1s. Expected time complexity is O(n).
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.
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.
int findSubArray(int arr[], int n){ int maxsize = -1, startindex; // variables to store result values // Create an auxiliary array sunmleft[]. sumleft[i] will be sum of array // elements from arr[0] to arr[i] 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;}Method 1
A simple method is to use two nested loops. The outer loop picks a starting point i. The inner loop considers all subarrays starting from i. If size of a subarray is greater than maximum size so far, then update the maximum size.
In the below code, 0s are considered as -1 and sum of all values from i to j is calculated. If sum becomes 0, then size of this subarray is compared with largest size so far.
int findSubArray(int arr[], int n){ int sum = 0; int maxsize = -1, startindex; // Pick a starting point as i for (int i = 0; i < n-1; i++) { sum = (arr[i] == 0)? -1 : 1; // Consider all subarrays starting from i for (int j = i+1; j < n; j++) { (arr[j] == 0)? sum += -1: sum += 1; // If this is a 0 sum subarray, then compare it with // maximum size subarray calculated so far if(sum == 0 && maxsize < j-i+1) { maxsize = j - i + 1; startindex = i; } } } if ( maxsize == -1 ) printf("No such subarray"); else printf("%d to %d", startindex, startindex+maxsize-1); return maxsize;}
No comments:
Post a Comment