Sunday, November 13, 2016

LeetCode 456 - 132 Pattern


http://bookshadow.com/weblog/2016/11/13/leetcode-132-pattern/
Given a sequence of n integers a1, a2, ..., an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.
Note: n will be less than 15,000.
Example 1:
Input: [1, 2, 3, 4]

Output: False

Explanation: There is no 132 pattern in the sequence.
Example 2:
Input: [3, 1, 4, 2]

Output: True

Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
Example 3:
Input: [-1, 3, 2, 0]

Output: True

Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
BST(Binary Search Tree 二叉查找树)
首先利用TreeMap(采用红黑树实现)统计nums中所有元素的个数,记为tm

变量min记录访问过元素的最小值

遍历数组nums,记当前数字为num

  将num在tm中的计数-1,计数为0时将num从tm中删去

  如果num < min,则更新min值为num

  否则,若tm中大于min的最小元素<num,则返回true

遍历结束,返回false
public boolean find132pattern(int[] nums) { TreeMap<Integer, Integer> tm = new TreeMap<>(); for (int num : nums) { tm.put(num, tm.getOrDefault(num, 0) + 1); } int min = Integer.MAX_VALUE; for (int num : nums) { int cnt = tm.get(num); if (cnt > 1) { tm.put(num, cnt - 1); } else { tm.remove(num); } if (num <= min) { min = num; } else { Integer target = tm.higherKey(min); if (target != null && target < num) { return true; } } } return false; }

X. Use Stack
https://discuss.leetcode.com/topic/68193/java-o-n-solution-using-stack-in-detail-explanation
The idea is that we can use a stack to keep track of previous min-max intervals.
Here is the principle to maintain the stack:
For each number num in the array
If stack is empty:
  • push a new Pair of num into stack
If stack is not empty:
  • if num < stack.peek().min, push a new Pair of num into stack
  • if num > stack.peek().min, we first pop() out the peek element, denoted as last
    • if num < last.max, we are done, return true;
    • if num > last.max, we merge num into last, which means last.max = num.
      Once we update last, if stack is empty, we just push back last.
      However, the crucial part is:
      If stack is not empty, the updated last might:
      • Entirely covered stack.peek(), i.e. last.min < stack.peek().min (which is always true) && last.max > stack.peek().max, in which case we keep popping out stack.peek().
      • Form a 1-3-2 pattern, we are done ,return true
So at any time in the stack, non-overlapping Pairs are formed in descending order by their min value, which means the min value of peek element in the stack is always the min value globally.
   class Pair{
        int min, max;
        public Pair(int min, int max){
            this.min = min;
            this.max = max;
        }
    }
    public boolean find132pattern(int[] nums) {
        Stack<Pair> stack = new Stack();
        for(int n: nums){
            if(stack.isEmpty() || n <stack.peek().min ) stack.push(new Pair(n,n));
            else if(n > stack.peek().min){ 
                Pair last = stack.pop();
                if(n < last.max) return true;
                else {
                    last.max = n;
                    while(!stack.isEmpty() && n >= stack.peek().max) stack.pop();
                    // At this time, n < stack.peek().max (if stack not empty)
                    if(!stack.isEmpty() && stack.peek().min < n) return true;
                    stack.push(last);
                }
                
            }
        }
        return false;
    }
https://discuss.leetcode.com/topic/67881/single-pass-c-o-n-space-and-time-solution-8-lines-with-detailed-explanation
We want to search for a sub sequence (s1,s2,s3)
INTUITION: The problem would be simple if we want to find sequence with s1 > s2 > s3, we just need to find s1, followed by s2 and s3. Now if we want to find a 132 sequence, we need to switch up the order of searching. we want to first find s2, followed by s3, then s1.
IDEA: We can start from either side but I think starting from the end allow us to finish in a single pass. The idea is to start from end and search for a candidate for s2 and s3. A number becomes a candidate for s3 if there is any number on the left (s2) that is bigger than it.
DETECTION: Keep track of the largest candidate of s3 and once we encounter any number smaller than s3, we know we found a valid sequence since s1 < s3 implies s1 < s2.
IMPLEMENTATION:
  1. Have a stack, each time we store a new number, we first pop out all numbers that are smaller than that number. The numbers that are popped out becomes candidate for s3.
  2. We keep track of the maximum of such s3.
  3. Once we encounter any number smaller than s3, we know we found a valid sequence since s1 < s3 implies s1 < s2.
RUNTIME: Each item is pushed and popped once at most, the time complexity is therefore O(n).
    bool find132pattern(vector<int>& nums) {
        int s3 = INT_MIN;
        stack<int> st;
        for( int i = nums.size()-1; i >= 0; i -- ){
            if( nums[i] < s3 ) return true;
            else while( !st.empty() && nums[i] > st.top() ){ 
              s3 = max( s3, st.top() ); st.pop(); 
            }
            st.push(nums[i]);
        }
        return false;
    }
public boolean find132pattern(int[] nums) {
    Stack<Integer> stack = new Stack<>();
    for (int i = nums.length - 1, two = Integer.MIN_VALUE; i >= 0; stack.push(nums[i--]))
        if (nums[i] < two) return true;
        else for (; !stack.empty() && nums[i] > stack.peek(); two = Math.max(two, stack.pop()));
    return false;
}
https://segmentfault.com/a/1190000007507137
维护一个pair, 里面有最大值和最小值。如果当前值小于pair的最小值,那么就将原来的pair压进去栈,然后在用这个新的pair的值再进行更新。如果当前值大于pair的最大值,首先这个值和原来在stack里面的那些pair进行比较,如果这个值比stack里面的值的max要大,就需要pop掉这个pair。如果没有适合返回的值,就重新更新当前的pair。
Class Pair {
    int min;
    int max;
    public Pair(int min, int max) {
        this.min = min;
        this.max = max;
    }
}

public boolean find123Pattern(int[] nums) {
    if(nums == null || nums.length < 3) return false;
    Pair cur = new Pair(nums[0], nums[0]);
    Stack<Pair> stack = new Stack<>();
    for(int i = 1; i < nums.length; i ++) {
        if(nums[i] < cur.min) {
            stack.push(cur);
            cur = new Pair(nums[i], nums[i]);
        }
        else if(nums[i] > cur.max) {
            while(!stack.isEmpty() && stack.peek().max <= nums[i]) {
                stack.pop();
            }
            if(!stack.isEmpty() && stack.peek.max > nums[i]) {
                return true;
            }
            cur.max = nums[i];
        }
        else if(nums[i] > cur.min && nums[i] < cur.max) {
            return true;
        }
    }
    return false;
}
X.
https://discuss.leetcode.com/topic/67615/share-my-easy-and-simple-solution
the worst case time complexity for this one is O(n^2) when the array is in ascending order, right?
Idea: Find peak and bottom
For every [bottom, peak], find if there is one number bottom<number<peak.
    public boolean find132pattern(int[] nums) {
        if(nums.length<3) return false;
        Integer low = null, high = null;
        int start = 0, end = 0;
        while(start<nums.length-1){
            while(start<nums.length-1 && nums[start]>=nums[start+1]) start++;
            // start is lowest now
            int m = start+1; //no need to use m - use end instead
            while(m<nums.length-1 && nums[m]<=nums[m+1]) m++;
            // m is highest now
            int j = m+1;
            while(j<nums.length){
                if(nums[j]>nums[start] && nums[j]<nums[m]) return true;
                j++;
            }
            start = m+1;
        }
        return false;
    }
X.
https://discuss.leetcode.com/topic/67592/java-straightforward
public boolean find132pattern(int[] nums) {
        int len = nums.length;
        if(len < 3) return false;
        int[] max_cache = new int[len];
        int[] min_cache = new int[len];
        
        min_cache[0] = nums[0];
        for(int j = 1; j < len ; j++){
            min_cache[j] = Math.min(min_cache[j-1], nums[j]);
        }
        max_cache[len-1] = nums[len-1];
        for(int j = len-2; j >= 0; j--){
            max_cache[j] = Math.max(max_cache[j+1], nums[j]);
        }
        for(int i = 1; i < len-1; i++){
            int val = nums[i];
            int left = min_cache[i-1];
            if(max_cache[i+1] > left && val > max_cache[i+1]) return true;
            for(int j = i+1; j < len; j++){
                if(nums[j] > left && val > nums[j]) return true;
            }
        }
        return false;
    }


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