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1663. Smallest String With A Given Numeric Value

Description

The numeric value of a lowercase character is defined as its position (1-indexed) in the alphabet, so the numeric value of a is 1, the numeric value of b is 2, the numeric value of c is 3, and so on.

The numeric value of a string consisting of lowercase characters is defined as the sum of its characters' numeric values. For example, the numeric value of the string "abe" is equal to 1 + 2 + 5 = 8.

You are given two integers n and k. Return the lexicographically smallest string with length equal to n and numeric value equal to k.

Note that a string x is lexicographically smaller than string y if x comes before y in dictionary order, that is, either x is a prefix of y, or if i is the first position such that x[i] != y[i], then x[i] comes before y[i] in alphabetic order.

 

Example 1:

Input: n = 3, k = 27
Output: "aay"
Explanation: The numeric value of the string is 1 + 1 + 25 = 27, and it is the smallest string with such a value and length equal to 3.

Example 2:

Input: n = 5, k = 73
Output: "aaszz"

 

Constraints:

  • 1 <= n <= 105
  • n <= k <= 26 * n

Solutions

  • class Solution {
    public String getSmallestString(int n, int k) {
        char[] ans = new char[n];
        Arrays.fill(ans, 'a');
        int i = n - 1, d = k - n;
        for (; d > 25; d -= 25) {
            ans[i--] = 'z';
        }
        ans[i] = (char) ('a' + d);
        return String.valueOf(ans);
    }
}

  • class Solution {
public:
    string getSmallestString(int n, int k) {
        string ans(n, 'a');
        int i = n - 1, d = k - n;
        for (; d > 25; d -= 25) {
            ans[i--] = 'z';
        }
        ans[i] += d;
        return ans;
    }
};

  • class Solution:
    def getSmallestString(self, n: int, k: int) -> str:
        ans = ['a'] * n
        i, d = n - 1, k - n
        while d > 25:
            ans[i] = 'z'
            d -= 25
            i -= 1
        ans[i] = chr(ord(ans[i]) + d)
        return ''.join(ans)


  • func getSmallestString(n int, k int) string {
	ans := make([]byte, n)
	for i := range ans {
		ans[i] = 'a'
	}
	i, d := n-1, k-n
	for ; d > 25; i, d = i-1, d-25 {
		ans[i] = 'z'
	}
	ans[i] += byte(d)
	return string(ans)
}

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