求和 < k 的 n 个连续数的最大和

Question

I am given an array and asked to find the maximum possible sum of n consecutive numbers, where the maximum sum is less than a given value k . For example:

array = {1, 3, 1, 2, 3, 4, 1}
k = 7

Here, the answer must be 6 , because the max sum that we can obtain that is less than 7 is: arr[1] + arr[2] + arr[3] = 3 + 1 + 2 = 6 How can I write an algorithm to find such a value?

(I have done it with a nested for loop, but it takes too much time, is there any other way to make this program work?)

Answer 1

Basic Foundation

First off, I'd suggest you read up a bit more about time complexity. There are enough good resources out there, and Complexity Theory is one of them. This should help you understand why your solution is not fast enough.

O(n^3) Solution

A brute-force approach is to check all possible subarrays, by iterating over the start and end points of the subarray, and adding up all elements in between.

Given an array arr of size n , the way to do that would be as follows:

for (int l = 0; l < n; ++l) {
    for (int r = l; r < n; ++r) {
        long sum = 0;
        for (int pos = l; pos <= r; ++pos) {
            sum += arr[pos];
        }
        if (sum < k)
            max = Math.max(max, sum);
    }
}

The final answer is stored in max .

O(n^2) Solution

A faster solution would eliminate the third loop by making use of prefix sums . This can be done as follows, with the help of an auxiliary array preSum of the same size as the main array, where preSum[i] stores the sum of the first i elements:

preSum[0] = arr[0];
for (int i = 1; i < n; ++i)
    preSum[i] = preSum[i - 1] + arr[i];
for (int l = 0; l < n; ++l) {
    for (int r = l; r < n; ++r) {
        long sum = preSum[r];
        if (l > 0)
            sum -= preSum[l - 1];
        if (sum < k)
            max = Math.max(max, sum);
    }
}

O(n) solution

The most efficient solution to this problem uses a sliding window / two-pointer approach. Note that we assume that negative numbers are not allowed.

We start with both l and r at the beginning of the array. There are two possible cases at every stage:

1. Sum of the current subarray <k : We can be hopeful and try to add more elements to the subarray. We do this by moving r one step further to the right.
2. Sum of the current subarray >=k : We need to remove some elements to make the sum satisfy the given constraint. This can be done by moving l one step to the right.

This is repeated till we hit we need to increment r , but have reached the end of the array. The code looks something like this:

long max = 0;
int l = 0;
int r = 0;
long sum = arr[0];
    while (true) {
    if (sum >= k) {
        sum -= arr[l];
        ++l;
    } else {
        if (r == n - 1)
            break;
        else {
            ++r;
            sum += arr[r];
        }
    }
    if (sum < k)
        max = Math.max(max, sum);
}