Kth Missing Positive Number

Given an array arr of positive integers sorted in a strictly increasing order, and an integer k.

Find the kth positive integer that is missing from this array.

Example

Input: arr=[2,3,4,7,11], k=5

Output: 9

Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.


Solution

Python solution

def binary_search(arr: List[int], k: int) -> int: if arr[0] > k: return k

1l = 0
2r = len(arr) - 1
3while l < r:
4    mid = (l + r) // 2
5    x = arr[mid] if mid < len(arr) else 10**9
6    if x - mid - 1 >= k:
7        r = mid
8    else:
9        l = mid + 1
10
11return k - (arr[l - 1] - (l - 1) - 1) + arr[l - 1]

Java Solution

public int binarySearch(int[] arr) { if (arr[0] > k) { return k; }

1int l = 0;
2int r = arr.length - 1;
3while (l < r) {
4    int mid = (l + r) / 2;
5    int x = mid < arr.length ? arr[mid] : Integer.MAX_VALUE;
6    if (x - mid - 1 >= k) {
7        r = mid;
8    } else {
9        l = mid + 1;
10    }
11}
12
13return k - (arr[l - 1] - (l - 1) - 1) + arr[l - 1];

}

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Solution Implementation

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What's the output of running the following function using input 56?

1KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12def letter_combinations_of_phone_number(digits):
13    def dfs(path, res):
14        if len(path) == len(digits):
15            res.append(''.join(path))
16            return
17
18        next_number = digits[len(path)]
19        for letter in KEYBOARD[next_number]:
20            path.append(letter)
21            dfs(path, res)
22            path.pop()
23
24    res = []
25    dfs([], res)
26    return res
27
1private static final Map<Character, char[]> KEYBOARD = Map.of(
2    '2', "abc".toCharArray(),
3    '3', "def".toCharArray(),
4    '4', "ghi".toCharArray(),
5    '5', "jkl".toCharArray(),
6    '6', "mno".toCharArray(),
7    '7', "pqrs".toCharArray(),
8    '8', "tuv".toCharArray(),
9    '9', "wxyz".toCharArray()
10);
11
12public static List<String> letterCombinationsOfPhoneNumber(String digits) {
13    List<String> res = new ArrayList<>();
14    dfs(new StringBuilder(), res, digits.toCharArray());
15    return res;
16}
17
18private static void dfs(StringBuilder path, List<String> res, char[] digits) {
19    if (path.length() == digits.length) {
20        res.add(path.toString());
21        return;
22    }
23    char next_digit = digits[path.length()];
24    for (char letter : KEYBOARD.get(next_digit)) {
25        path.append(letter);
26        dfs(path, res, digits);
27        path.deleteCharAt(path.length() - 1);
28    }
29}
30
1const KEYBOARD = {
2    '2': 'abc',
3    '3': 'def',
4    '4': 'ghi',
5    '5': 'jkl',
6    '6': 'mno',
7    '7': 'pqrs',
8    '8': 'tuv',
9    '9': 'wxyz',
10}
11
12function letter_combinations_of_phone_number(digits) {
13    let res = [];
14    dfs(digits, [], res);
15    return res;
16}
17
18function dfs(digits, path, res) {
19    if (path.length === digits.length) {
20        res.push(path.join(''));
21        return;
22    }
23    let next_number = digits.charAt(path.length);
24    for (let letter of KEYBOARD[next_number]) {
25        path.push(letter);
26        dfs(digits, path, res);
27        path.pop();
28    }
29}
30

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