LeetCode 2515. Shortest Distance to Target String in a Circular Array Solution in Java, C++, Python & More | Explanation + Code

Description

You are given a 0-indexed circular string array words and a string target. A circular array means that the array's end connects to the array's beginning.

• Formally, the next element of words[i] is words[(i + 1) % n] and the previous element of words[i] is words[(i - 1 + n) % n], where n is the length of words.

Starting from startIndex, you can move to either the next word or the previous word with 1 step at a time.

Return the shortest distance needed to reach the string target. If the string target does not exist in words, return -1.

 

Example 1:

Input: words = ["hello","i","am","leetcode","hello"], target = "hello", startIndex = 1
Output: 1
Explanation: We start from index 1 and can reach "hello" by
- moving 3 units to the right to reach index 4.
- moving 2 units to the left to reach index 4.
- moving 4 units to the right to reach index 0.
- moving 1 unit to the left to reach index 0.
The shortest distance to reach "hello" is 1.

Example 2:

Input: words = ["a","b","leetcode"], target = "leetcode", startIndex = 0
Output: 1
Explanation: We start from index 0 and can reach "leetcode" by
- moving 2 units to the right to reach index 2.
- moving 1 unit to the left to reach index 2.
The shortest distance to reach "leetcode" is 1.

Example 3:

Input: words = ["i","eat","leetcode"], target = "ate", startIndex = 0
Output: -1
Explanation: Since "ate" does not exist in words, we return -1.

 

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 100
• words[i] and target consist of only lowercase English letters.
• 0 <= startIndex < words.length

Solutions

Solution 1: Single Traversal

We traverse the array words, find the words equal to target, and compute their distance t from startIndex. The shortest distance in this case is min(t, n - t), so we only need to keep updating the minimum value.

The time complexity is O(n), where n is the length of the array. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustC

class Solution: def closestTarget(self, words: List[str], target: str, startIndex: int) -> int: n = len(words) ans = n for i, w in enumerate(words): if w == target: t = abs(i - startIndex) ans = min(ans, t, n - t) return -1 if ans == n else ans(code-box)

class Solution { public int closestTarget(String[] words, String target, int startIndex) { int n = words.length; int ans = n; for (int i = 0; i < n; i++) { if (words[i].equals(target)) { int t = Math.abs(i - startIndex); ans = Math.min(ans, Math.min(t, n - t)); } } return ans == n ? -1 : ans; } }(code-box)

class Solution { public: int closestTarget(vector<string>& words, string target, int startIndex) { int n = words.size(); int ans = n; for (int i = 0; i < n; i++) { if (words[i] == target) { int t = abs(i - startIndex); ans = min({ans, t, n - t}); } } return ans == n ? -1 : ans; } };(code-box)

func closestTarget(words []string, target string, startIndex int) int { n := len(words) ans := n for i, w := range words { if w == target { t := i - startIndex if t < 0 { t = -t } ans = min(ans, t, n-t) } } if ans == n { return -1 } return ans }(code-box)

function closestTarget(words: string[], target: string, startIndex: number): number { const n = words.length; let ans = n; for (let i = 0; i < n; i++) { if (words[i] === target) { const t = Math.abs(i - startIndex); ans = Math.min(ans, t, n - t); } } return ans === n ? -1 : ans; }(code-box)

impl Solution { pub fn closest_target(words: Vec<String>, target: String, start_index: i32) -> i32 { let n = words.len() as i32; let mut ans = n; for (i, w) in words.iter().enumerate() { if w == &target { let t = (i as i32 - start_index).abs(); ans = ans.min(t.min(n - t)); } } if ans == n { -1 } else { ans } } }(code-box)

#define min(a, b) ((a) < (b) ? (a) : (b)) int closestTarget(char** words, int wordsSize, char* target, int startIndex) { int n = wordsSize; int ans = n; for (int i = 0; i < n; i++) { if (strcmp(words[i], target) == 0) { int t = abs(i - startIndex); int dist = min(t, n - t); ans = min(ans, dist); } } return ans == n ? -1 : ans; }(code-box)