1633 - Smallest String With A Given Numeric Value (Medium)

Problem Link

https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/

Problem Statement

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 <= 10^5
• n <= k <= 26 * n

Approach 1: Build from the right

To obtain lexicographically smallest string, we should put $a$ from the left and $z$ from the right if possible and put what's left in the middle. Therefore, we initialise the answer with all $a$s. Starting from the right, the best case is to make it to $z$ (i.e. $s[i] + 25$). If we cannot do it, then we can only make it to the max one (i.e. $s[i] + k$).

Written by @wingkwong

class Solution {
public:
    string getSmallestString(int n, int k) {
        // init s with all 'a's
        string s(n, 'a');
        // since we put n times, update k
        k -= n; 
        // 0-index
        n -= 1;
        // fill character until k <= 0
        while (k > 0) {
            // we either put a + 25 (z) or a + k
            s[n] += min(k, 25);
            // update k
            k -= 25;
            // shift the pointer to the left
            n -= 1;
        }
        return answer
        return s;
    }
};

Approach 2: Three segments

The answer can be potentially constructed by three segments. The first segment contains only $a$. The last segment contains only $z$. The middle segment contains a character. This approach is to calculate each character in each segment given $n$ and $k$.

First we calculate the number of characters that are not $a$, i.e. the total number of characters in the second and the third segment. Let's call it $nonA$. Then we know that there would be $n - nonA$ a in the first segment and $nonA - 1$ z in the third segment (minus one because we need one for the second segment). So how to get $nonA$? We can use above condition to find out $nonA$. That is $(n - nonA) * 1 + nonA * 26 >= k$ which gives $nonA >= (k - n) / 25$.

For the middle segment, how many $k$ left we can use? We've used $(n - nonA) * 1$ for the first segment and $(nonA-1)* 26$ for the last segment. The index of the character in the middle segment would be $k - (n - nonA) - (nonA - 1) * 26$.

Written by @wingkwong

class Solution {
public:
    string getSmallestString(int n, int k) {
        // total number of characters in 2nd & 3rd segment
        int nonA = ((k - n) + 25 - 1) / 25;
        // 1st segment: there is (n - nonA) 'a's
        // 2nd segment: one character with index k - (n - nonA) - (nonA - 1) * 26
        // 3rd segemnt: there is (nonA - 1) 'z's
        return string(n - nonA, 'a') + 
               char(k - (n - nonA) - (nonA - 1) * 26 + 'a' - 1) + 
               string(nonA - 1, 'z');
    }
};