LeetCode 525 - Contiguous Array

https://leetcode.com/problems/contiguous-array

Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.

Example 1:

Input: [0,1]
Output: 2
Explanation: [0, 1] is the longest contiguous subarray with equal number of 0 and 1.

Example 2:

Input: [0,1,0]
Output: 2
Explanation: [0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.

Note: The length of the given binary array will not exceed 50,000.

X.
https://discuss.leetcode.com/topic/80020/one-pass-use-a-hashmap-to-record-0-1-count-difference

nice solution without modifying current array content.

    public int findMaxLength(int[] nums) {
        HashMap<Integer,Integer> map=new HashMap<>();
        map.put(0,-1);
        
        int zero=0;
        int one=0;
        int len=0;
        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){
                zero++;
            }else{
                one++;
            }
            
            if(map.containsKey(zero-one)){
                len=Math.max(len,i-map.get(zero-one));
            }else{
                map.put(zero-one,i);
            }
        }
        
        return len;
    }

https://leetcode.com/articles/contiguous-array/

    public int findMaxLength(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int maxlen = 0, count = 0;
        for (int i = 0; i < nums.length; i++) {
            count = count + (nums[i] == 1 ? 1 : -1);
            if (map.containsKey(count)) {
                maxlen = Math.max(maxlen, i - map.get(count));
            } else {
                map.put(count, i);
            }
        }
        return maxlen;
    }

https://discuss.leetcode.com/topic/79906/easy-java-o-n-solution-presum-hashmap

The idea is to change 0 in the original array to -1. Thus, if we find SUM[i, j] == 0 then we know there are even number of -1 and 1 between index i and j. Also put the sum to index mapping to a HashMap to make search faster.

sumToIndex is a hash table stores the accumulative total's corresponding index. If and only if we find two different indices i, j and the two corresponding sum sumToIndex[sum]for i, j are equal, we claim that sum(nums[j+1:i+1]) == 0.

Consider sumToIndex.get(sum) as j, we wanna update answer with j - i which is exactly i - sumToIndex.get(sum).

    public int findMaxLength(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) nums[i] = -1;
        }
        
        Map<Integer, Integer> sumToIndex = new HashMap<>();
        sumToIndex.put(0, -1);
        int sum = 0, max = 0;
        
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (sumToIndex.containsKey(sum)) {
                max = Math.max(max, i - sumToIndex.get(sum));
            }
            else {
                sumToIndex.put(sum, i);
            }
        }
        
        return max;
    }

I had a similar idea, but in one pass:

    public int findMaxLength(int[] nums) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>() {{put(0,0);}};
        int maxLength = 0, runningSum = 0;
        for (int i=0;i<nums.length;i++) {
            runningSum += nums[i];
            Integer prev = map.get(2*runningSum-i-1);//?
            if (prev != null) maxLength = Math.max(maxLength, i+1-prev);
            else map.put(2*runningSum-i-1, i+1);
        }
        return maxLength;
    }

https://discuss.leetcode.com/topic/79977/python-and-java-with-little-tricks-incl-a-oneliner/2

Using putIfAbsent so I only need one map function call per number.

public int findMaxLength(int[] nums) {
    Map<Integer, Integer> index = new HashMap<>();
    index.put(0, -1);
    int balance = 0, maxlen = 0;
    for (int i = 0; i < nums.length; i++) {
        balance += nums[i] * 2 - 1;
        Integer first = index.putIfAbsent(balance, i);
        if (first != null)
            maxlen = Math.max(maxlen, i - first);
    }
    return maxlen;
}

Could avoid using Math.max like this:

        if (first != null && i - first > maxlen)
            maxlen = i - first;

https://discuss.leetcode.com/topic/79932/java-one-pass-o-n-solution-with-explanation

diff[i] is "count of 1s" minus "count of 0s" so far.

For given i and j, if diff[i] == diff[j], then the subarray between i and j is a contiguous array as defined in the questions.

For any value of diff[], we save the index of the first item in the hashmap. Then once the value appears again, we get the length of the subarray between the current index and the index of the first item.

public int findMaxLength(int[] nums) {
    int res = 0;
    int n = nums.length;
    
    int[] diff = new int[n + 1];
    
    Map<Integer, Integer> map = new HashMap<>();
    map.put(0, 0);
    
    for (int i = 1; i <= n; i++) {
        diff[i] = diff[i - 1] + (nums[i - 1] == 0 ? -1 : 1);

        if (!map.containsKey(diff[i]))
            map.put(diff[i], i);
        else
            res = Math.max(res, i - map.get(diff[i]));
    }

    return res;
}

https://discuss.leetcode.com/topic/80056/python-o-n-solution-with-visual-explanation/2

    def findMaxLength(self, nums):
        count = 0
        max_length=0
        table = {0: 0}
        for index, num in enumerate(nums, 1):
            if num == 0:
                count -= 1
            else:
                count += 1
            
            if count in table:
                max_length = max(max_length, index - table[count])
            else:
                table[count] = index
        
        return max_length

X.
https://discuss.leetcode.com/topic/79907/share-my-dp-map-solution-one-pass/2
Use dp[][] to store the zeros and ones from start to current position
and use map to store the difference of ones and zeros with index;
if current position ones-zeros = 2, we check map if there is other position ones-zeros=2
if we find, the 2*(current index - that index) is candidate
the max candidate is the answer

public int findMaxLength(int[] nums) {
 int n = nums.length, res = 0;
 Map<Integer, Integer> map = new HashMap<>();
 int[][] dp = new int[n+1][2];
 for (int i = 1; i < dp.length; i++) {
  if (nums[i-1] == 0) {
   dp[i][0] = dp[i-1][0]+1;
   dp[i][1] = dp[i-1][1];
  }else {
   dp[i][0] = dp[i-1][0];
   dp[i][1] = dp[i-1][1]+1;
  }
  if (dp[i][0] == dp[i][1]) res = Math.max(res, dp[i][0]*2);
  else {
   int dif = dp[i][1]-dp[i][0];
   if (map.containsKey(dif)) res = Math.max(res, 2*(dp[i][0]-dp[map.get(dif)][0]));
   else map.put(dif, i);
  }
 }
 return res;
}

X.
https://leetcode.com/articles/contiguous-array/

The brute force approach is really simple. We consider every possible subarray within the given array and count the number of zeros and ones in each subarray. Then, we find out the maximum size subarray with equal no. of zeros and ones out of them.

    public int findMaxLength(int[] nums) {
        int maxlen = 0;
        for (int start = 0; start < nums.length; start++) {
            int zeroes = 0, ones = 0;
            for (int end = start; end < nums.length; end++) {
                if (nums[end] == 0) {
                    zeroes++;
                } else {
                    ones++;
                }
                if (zeroes == ones) {
                    maxlen = Math.max(maxlen, end - start + 1);
                }
            }
        }
        return maxlen;
    }