Leetcode327: Count of Range Sum 范围和个数问题

问题描述

给定一个整数数组，返回range sum 落在给定区间[lower, upper] (包含lower和upper)的个数。range sum S(i, j) 表示数组中第i 个元素到j 个元素之和。

Note:
A naive algorithm of O(n2) is trivial. You MUST do better than that.

Example:

Input: nums = [-2,5,-1], lower = -2, upper = 2,
Output: 3
Explanation: The three ranges are : [0,0], [2,2], [0,2] and their respective sums are: -2, -1, 2.

分析

这个题目比较难，楼主第一次面对这种题型，直接缴械投降。参考了各位大神的解题思路，总结了两种解法。一种是TreeMap思路，另外一种是使用segment tree (or binary index tree)。题目寻找需要range sum 在[lower, upper] 之间的个数，满足条件的case用数学公式表达为：

lower <= sum[i] - sum[j] <= upper, i > j, sum[i] is prefix sum of nums at index of i.

也就是

sum[i] - high <=  sum[j] <= sum[i] - lower, i > j, sum[i] is prefix sum of nums at index of i.

(or

lower + sum[j] <=  sum[i] <= sum[j] + higher, i > j, sum[i] is prefix sum of nums at index of i.)

那么我们的问题可以转化为求落在[sum[i] - high,sum[i] - lower] 区间sum[j]的个数, i = 0....n, j < i。

无论是TreeMap还是Segment Tree，总体的时间复杂度都为nlogn。

实现

TreeMap

TreeMap 的key 是prefixsum, value 是相对应的个数。主要使用TreeMap的subMap的方法，求得落在区间内[sum[i] - high, sum[i] - lower]的sum[j]的个数。

 public int countRangeSum(int[] nums, int lower, int upper) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        //key is the sum[i], value is the corresponding count
        // sum[i] - sum[j] in [lower, upper], transform to find how many sum[j] 在区间[sum[i] - high, sum[i] - lower]。
        TreeMap<Long, Integer> map = new TreeMap();
        long sum = 0;
        int cnt = 0;
        
        for(int i = 0; i < nums.length; i++){
            sum += nums[i];
            //sum[0, i]满足case
            if(sum >= lower && sum <= upper){
                cnt++;
            }
            //find sum[j] 的个数that lies in [sum[i] - high, sum[i] - lower]之间
            cnt += map.subMap(sum - upper, true, sum - lower, true).values().stream().mapToInt(Integer::valueOf).sum();
             
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
        return cnt;
    }

Segment Tree

Segment Tree每个节点保存区间段的范围和落在这个区间内prefix sum的个数。

  class Node {
        Node left;
        Node right;
        //落在区间内的个数
        int count;
        long min;
        long max;
        public Node(long min, long max) {
            this.min = min;
            this.max = max;
        }
    }
   //构建segement tree
    private Node buildTree(Long[] valArr, int low, int high) {
        if(low > high) return null;
        Node root = new Node(valArr[low], valArr[high]);
        if(low == high) return root;
        int mid = low + (high - low)/2;
        root.left = buildTree(valArr, low, mid);
        root.right = buildTree(valArr, mid+1, high);
        return root;
    }
    
    private void update(Node root, Long val) {
        if(root == null) return;
        if(val >= root.min && val <= root.max) {
            root.count++;
            update(root.left, val);
            update(root.right, val);
        }
    }
    
    private int query(Node root, long min, long max) {
        if(root == null) return 0;
        if(min > root.max || max < root.min) return 0;
        if(min <= root.min && max >= root.max) return root.count;
        return query(root.left, min, max) + query(root.right, min, max);
    }

    public int countRangeSum(int[] nums, int lower, int upper) {

        if(nums == null || nums.length == 0) return 0;
        int ans = 0;
        Set<Long> valSet = new HashSet<Long>();
        long sum = 0;
        for(int i = 0; i < nums.length; i++) {
            sum += (long) nums[i];
            valSet.add(sum);
        }

        Long[] valArr = valSet.toArray(new Long[0]);

        Arrays.sort(valArr);
        Node root = buildTree(valArr, 0, valArr.length-1);
        
        sum = nums[0];
        ans += (sum >= lower && sum <= upper) ? 1:0; 
        for(int i = 1; i < nums.length; i++) {
            //sum[i]
            update(root, sum);
            //sum[j]
            sum += (long) nums[i];
            ans += (sum >= lower && sum <= upper) ? 1:0; 
            ans += query(root, (long)sum - upper, (long)sum - lower);
        }
        return ans;
    }