1663. Smallest String With A Given Numeric Value

1663. Smallest String With A Given Numeric Value (Medium)

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

Companies:
Lendingkart

Related Topics:
String, Greedy

Solution 1. Greedy

Intuition: We can do it greedily:

• If picking a won't result in unsolvable problem, we prepend a to the start of the string.
• Otherwise, if picking z won't result in unsolvable problem, we append z to the end of the string.
• Otherwise, if there is still a non-zero k value left (which must be smaller than 26), we add 'a' + k - 1 in the middle.

Algorithm:

How to check if it will become unsolvable problem?

If after picking 'a', the remainder numeric value k - 1 can't be formed even if using all 'z's, i.e. k - 1 >= (n - 1) * 26, then it's unsolvable.

So if n - 1 > 0 && k - 1 < (n - 1) * 26, we should keep prepending 'a'.

If after picking 'z', the remainder numeric value k - 26 can't be formed even if using all 'a's, i.e. k - 26 < n - 1, then it's unsolvable.

So if n - 1 > 0 && k - 26 >= n - 1, we should keep appending 'z'.

// OJ: https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string getSmallestString(int n, int k) {
        string s;
        int a = 0, z = 0;
        while (n - 1 > 0 && k - 1 < (n - 1) * 26) ++a, --k, --n;
        while (n - 1 > 0 && k - 26 >= n - 1) ++z, k -= 26, --n;
        while (a-- > 0) s += 'a';
        if (k) s += 'a' + k - 1;
        while (z-- > 0) s += 'z';
        return s;
    }
};

Solution 2. Greedy

Starts with all 'a', fill the k value from the end of the array backwards.

// OJ: https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string getSmallestString(int n, int k) {
        string s(n, 'a');
        k -= n;
        for (int i = n - 1; k > 0; --i) {
            int d = min(k, 25);
            s[i] += d;
            k -= d;
        }
        return s;
    }
};

Solution 3. Greedy

Initialize ans as all zs. Try filling a from the left. If we fill a at and before index i, we have k - i - 1 values left which should be <= 26 * (n - i - 1) i.e. the value of all the zs to the right of i. So, we keep filling a when k - i - 1 <= 26 * (n - i - 1).

When we exit the loop, ans[i] might not be z. The value left for ans[i] is k minus the values created by the as to the left, i.e. i, and minus the values created by the zs to the right, i.e. 26 * (n - i - 1). So, the value for ans[i] is k - i - (n - i - 1) * 26.

// OJ: https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    string getSmallestString(int n, int k) {
        string ans(n, 'z');
        int i = 0;
        for (; i < n && k - i - 1 <= 26 * (n - i - 1); ++i) {
            ans[i] = 'a';
        }
        ans[i] = 'a' + k - i - 1 - (n - i - 1) * 26;
        return ans;
    }
};