LintCode 1753: Doing Homework Algorithms Medium

1753 · Doing Homework Algorithms Medium
Description
For n people, each of them needs to do m jobs independently.
The i job takes cost[i] time. Since each person’s free time is different, the i person has val[i] time, which means that the total time for his jobs will not exceed val[i]. Everyone starts with the first job, then the 2nd, the 3rd… Now, you need to figure out how much time they spend.

1<=n<=100000
1<=m<=100000
1<=val[i]<=100000
1<=cost[i]<=100000

Example
Example 1:

Given cost=[1,2,3,5],val=[6,10,4], return 15.
Input:
[1,2,3,5]
[6,10,4]
Output:
15

Explanation:
The first person can complete the 1st job, the 2nd job, the 3rd job, 1+2+3<=6.
The second person cancomplete the 1st job, the 2nd job, the 3rd job, and cannot complete the 4th job, 1+2+3<=10, 1+2+3+5>10.
The third person can complete the 1st job, the 2nd job, and cannot complete the 3rd job, 1+2<=4, 1+2+3>4.
1+2+3+1+2+3+1+2=15, returning 15.
Example 2:

Given cost=[3,7,3,2,5],val=[10,20,12,8,17,25], return 78.
Input:
[3,7,3,2,5]
[10,20,12,8,17,25]
Output:
78

Explanation:
The first person can complete the 1st job, the 2nd job, 3 + 7<=10.
The second person cancomplete the 1st job, the 2nd job, the 3rd job, the 4th job,the 5th job, 3+7+3+2+5<=20
The third person can complete the 1st job, the 2nd job, and cannot complete the 3rd job, 3+7<=12,3+7+3>12.
The forth person can complete the 1st job, and cannot complete the 2nd job, 3<=8,7+3>8.
The fifth person can complete the 1st job, the 2nd job, the 3rd job, the 4th job, and cannot complete the 5th job,3+7+3+2<=17,3+7+3+2+5>17.
The sixth person cancomplete the 1st job, the 2nd job, the 3rd job, the 4th job,the 5th job, 3+7+3+2+5<=25
3+7+3+7+3+2+5+3+7+3+3+7+3+2+3+7+3+2+5=78, returning 78.

解法1：Presums + binary search

class Solution {
public:
    /**
     * @param cost: the cost
     * @param val: the val
     * @return: the all cost
     */
    long long doingHomework(vector<int> &cost, vector<int> &val) {
        int len_cost = cost.size();
        int len_val = val.size();
        vector<long long> preSums(len_cost, 0);
        preSums[0] = cost[0];
        for (int i = 1; i < len_cost; i++) {
            preSums[i] = preSums[i - 1] + cost[i];
        }
        long long res = 0;
        for (int i = 0; i < len_val; i++) {
            long long findIndex = binarySearch(preSums, val[i]);
            if (findIndex >= 0) res += preSums[findIndex];
        }
        return res;
    }
private:
    int binarySearch(vector<long long> &nums, int target) {
        int start = 0, end = nums.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] < target) {
                start = mid;
            } else if (nums[mid] > target) {
                end = mid;
            } else {
                return mid;
            }
        }
        if (nums[end] <= target) return end;
        if (nums[start] <= target) return start;
        return -1;
    }
};