LCP 12. 小张刷题计划 

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class Solution {
    public int minTime(int[] time, int m) {
        int sum = 0;
        for (int t : time) {
            sum += t;
        }

        // 搜索最小的 T
        int left = 0, right = sum; // [left, right]
        while (left < right) {
            int mid = (left + right) >>> 1;
            if (canFinish(time, mid, m)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }

        return left;
    }

    private boolean canFinish(int[] time, int t, int maxDays) {
        int subMax = 0, subSum = 0, subDays = 1;

        for (int num : time) {
            int secondMax = Math.min(subMax, num); // 第二大的耗时
            if (subSum + secondMax <= t) { // 每天删除最大值后小于等于 t
                subSum += secondMax;
                subMax = Math.max(subMax, num);
            } else { // 新的一天
                subSum = 0; // 默认将当天删除，所以此处为 0
                subDays++;
                subMax = num;
            }
        }

        return subDays <= maxDays;
    }
}

References

LCP 12. 小张刷题计划