Subarray Sum Closest

Problem

Given an integer array, find a subarray with sum closest to zero. Return the indexes of the first number and last number.

Example

Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4].

Analysis

相比于Subarray Sum问题,这里同样可以记录下位置i的sum,存入一个数组或者链表中,按照sum的值sort,再寻找相邻两个sum差值绝对值最小的那个,也就得到了subarray sum closest to 0。

延伸:Subsequence with sum closest to t

Solution

class Pair {
    int sum;
    int index;
    public Pair(int s, int i) {
        sum = s;
        index = i;
    }
}

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number
     *          and the index of the last number
     */
    public int[] subarraySumClosest(int[] nums) {
        int[] res = new int[2];
        if (nums == null || nums.length == 0) {
            return res;
        }

        int len = nums.length;
        if(len == 1) {
            res[0] = res[1] = 0;
            return res;
        }
        Pair[] sums = new Pair[len+1];
        int prev = 0;
        sums[0] = new Pair(0, 0);
        for (int i = 1; i <= len; i++) {
            sums[i] = new Pair(prev + nums[i-1], i);
            prev = sums[i].sum;
        }
        Arrays.sort(sums, new Comparator<Pair>() {
           public int compare(Pair a, Pair b) {
               return a.sum - b.sum;
           }
        });
        int ans = Integer.MAX_VALUE;
        for (int i = 1; i <= len; i++) {

            if (ans > sums[i].sum - sums[i-1].sum) {
                ans = sums[i].sum - sums[i-1].sum;
                int[] temp = new int[]{sums[i].index - 1, sums[i - 1].index - 1};
                Arrays.sort(temp);
                res[0] = temp[0] + 1;
                res[1] = temp[1];
            }
        }

        return res;
    }
}

Reference

http://www.jiuzhang.com/solutions/subarray-sum-closest/

http://rafal.io/posts/subsequence-closest-to-t.html

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