Formatted question description: https://leetcode.ca/all/1663.html

# 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

Related Topics:
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/
// 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.

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/
// 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;
}
};


Java

• class Solution {
public String getSmallestString(int n, int k) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
int remainLength = n - i - 1;
int remainSum = 26 * remainLength;
array[i] = Math.max(1, k - remainSum);
k -= array[i];
}
StringBuffer sb = new StringBuffer();
for (int i = 0; i < n; i++) {
char c = (char) ('a' + array[i] - 1);
sb.append(c);
}
return sb.toString();
}
}

• // OJ: https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/
// 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;
}
};

• # 1663. Smallest String With A Given Numeric Value
# https://leetcode.com/problems/smallest-string-with-a-given-numeric-value/

class Solution:
def getSmallestString(self, n: int, k: int) -> str:

res = []

while k > 0 and n > 0:
if (26 + n) <= k:
res.append('z')
k -= 26
n -= 1
else:
if n == 1:
t = k % 27
res.append(chr(t+96))
k -= t
n -= 1
else:
t = (k-n+1) % 27
res.append(chr(t+96))
k -= t
n -= 1

res.sort()

return "".join(res)