Question

209. Minimum Size Subarray Sum

中文文档

Description

Given an array of positive integers nums and a positive integer target, return _the minimal length of a __subarray__ whose sum is greater than or equal to_ target. If there is no such subarray, return 0 instead.

 

Example 1:

Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

 

Constraints:

• 1 <= target <= 109
• 1 <= nums.length <= 105
• 1 <= nums[i] <= 104

 

Follow up: If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log(n)).

Solutions

Solution 1: Prefix Sum + Binary Search

First, we preprocess the prefix sum array $s$ of the array $nums$, where $s[i]$ represents the sum of the first $i$ elements of the array $nums$. Since all elements in the array $nums$ are positive integers, the array $s$ is also monotonically increasing. Also, we initialize the answer $ans = n + 1$, where $n$ is the length of the array $nums$.

Next, we traverse the prefix sum array $s$. For each element $s[i]$, we can find the smallest index $j$ that satisfies $s[j] \geq s[i] + target$ by binary search. If $j \leq n$, it means that there exists a subarray that satisfies the condition, and we can update the answer, i.e., $ans = min(ans, j - i)$.

Finally, if $ans \leq n$, it means that there exists a subarray that satisfies the condition, return $ans$, otherwise return $0$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

class Solution:
  def minSubArrayLen(self, target: int, nums: List[int]) -> int:
    n = len(nums)
    s = list(accumulate(nums, initial=0))
    ans = n + 1
    for i, x in enumerate(s):
      j = bisect_left(s, x + target)
      if j <= n:
        ans = min(ans, j - i)
    return ans if ans <= n else 0

class Solution {
  public int minSubArrayLen(int target, int[] nums) {
    int n = nums.length;
    long[] s = new long[n + 1];
    for (int i = 0; i < n; ++i) {
      s[i + 1] = s[i] + nums[i];
    }
    int ans = n + 1;
    for (int i = 0; i <= n; ++i) {
      int j = search(s, s[i] + target);
      if (j <= n) {
        ans = Math.min(ans, j - i);
      }
    }
    return ans <= n ? ans : 0;
  }

  private int search(long[] nums, long x) {
    int l = 0, r = nums.length;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (nums[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
}

class Solution {
public:
  int minSubArrayLen(int target, vector<int>& nums) {
    int n = nums.size();
    vector<long long> s(n + 1);
    for (int i = 0; i < n; ++i) {
      s[i + 1] = s[i] + nums[i];
    }
    int ans = n + 1;
    for (int i = 0; i <= n; ++i) {
      int j = lower_bound(s.begin(), s.end(), s[i] + target) - s.begin();
      if (j <= n) {
        ans = min(ans, j - i);
      }
    }
    return ans <= n ? ans : 0;
  }
};

func minSubArrayLen(target int, nums []int) int {
  n := len(nums)
  s := make([]int, n+1)
  for i, x := range nums {
    s[i+1] = s[i] + x
  }
  ans := n + 1
  for i, x := range s {
    j := sort.SearchInts(s, x+target)
    if j <= n {
      ans = min(ans, j-i)
    }
  }
  if ans == n+1 {
    return 0
  }
  return ans
}

function minSubArrayLen(target: number, nums: number[]): number {
  const n = nums.length;
  const s: number[] = new Array(n + 1).fill(0);
  for (let i = 0; i < n; ++i) {
    s[i + 1] = s[i] + nums[i];
  }
  let ans = n + 1;
  const search = (x: number) => {
    let l = 0;
    let r = n + 1;
    while (l < r) {
      const mid = (l + r) >>> 1;
      if (s[mid] >= x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  };
  for (let i = 0; i <= n; ++i) {
    const j = search(s[i] + target);
    if (j <= n) {
      ans = Math.min(ans, j - i);
    }
  }
  return ans === n + 1 ? 0 : ans;
}

impl Solution {
  pub fn min_sub_array_len(target: i32, nums: Vec<i32>) -> i32 {
    let n = nums.len();
    let mut res = n + 1;
    let mut sum = 0;
    let mut i = 0;
    for j in 0..n {
      sum += nums[j];

      while sum >= target {
        res = res.min(j - i + 1);
        sum -= nums[i];
        i += 1;
      }
    }
    if res == n + 1 {
      return 0;
    }
    res as i32
  }
}

public class Solution {
  public int MinSubArrayLen(int target, int[] nums) {
    int n = nums.Length;
    long s = 0;
    int ans = n + 1;
    for (int i = 0, j = 0; i < n; ++i) {
      s += nums[i];
      while (s >= target) {
        ans = Math.Min(ans, i - j + 1);
        s -= nums[j++];
      }
    }
    return ans == n + 1 ? 0 : ans;
  }
}

Solution 2: Two Pointers

We can use two pointers $j$ and $i$ to maintain a window, where the sum of all elements in the window is less than $target$. Initially, $j = 0$, and the answer $ans = n + 1$, where $n$ is the length of the array $nums$.

Next, the pointer $i$ starts to move to the right from $0$, moving one step each time. We add the element corresponding to the pointer $i$ to the window and update the sum of the elements in the window. If the sum of the elements in the window is greater than or equal to $target$, it means that the current subarray satisfies the condition, and we can update the answer, i.e., $ans = \min(ans, i - j + 1)$. Then we continuously remove the element $nums[j]$ from the window until the sum of the elements in the window is less than $target$, and then repeat the above process.

Finally, if $ans \leq n$, it means that there exists a subarray that satisfies the condition, return $ans$, otherwise return $0$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $nums$.

class Solution:
  def minSubArrayLen(self, target: int, nums: List[int]) -> int:
    n = len(nums)
    ans = n + 1
    s = j = 0
    for i, x in enumerate(nums):
      s += x
      while j < n and s >= target:
        ans = min(ans, i - j + 1)
        s -= nums[j]
        j += 1
    return ans if ans <= n else 0

class Solution {
  public int minSubArrayLen(int target, int[] nums) {
    int n = nums.length;
    long s = 0;
    int ans = n + 1;
    for (int i = 0, j = 0; i < n; ++i) {
      s += nums[i];
      while (j < n && s >= target) {
        ans = Math.min(ans, i - j + 1);
        s -= nums[j++];
      }
    }
    return ans <= n ? ans : 0;
  }
}

class Solution {
public:
  int minSubArrayLen(int target, vector<int>& nums) {
    int n = nums.size();
    long long s = 0;
    int ans = n + 1;
    for (int i = 0, j = 0; i < n; ++i) {
      s += nums[i];
      while (j < n && s >= target) {
        ans = min(ans, i - j + 1);
        s -= nums[j++];
      }
    }
    return ans == n + 1 ? 0 : ans;
  }
};

func minSubArrayLen(target int, nums []int) int {
  n := len(nums)
  s := 0
  ans := n + 1
  for i, j := 0, 0; i < n; i++ {
    s += nums[i]
    for s >= target {
      ans = min(ans, i-j+1)
      s -= nums[j]
      j++
    }
  }
  if ans == n+1 {
    return 0
  }
  return ans
}

function minSubArrayLen(target: number, nums: number[]): number {
  const n = nums.length;
  let s = 0;
  let ans = n + 1;
  for (let i = 0, j = 0; i < n; ++i) {
    s += nums[i];
    while (s >= target) {
      ans = Math.min(ans, i - j + 1);
      s -= nums[j++];
    }
  }
  return ans === n + 1 ? 0 : ans;
}