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Formatted question description: https://leetcode.ca/all/1540.html

# 1540. Can Convert String in K Moves (Medium)

Given two strings s and t, your goal is to convert s into t in k moves or less.

During the ith (1 <= i <= kmove you can:

• Choose any index j (1-indexed) from s, such that 1 <= j <= s.length and j has not been chosen in any previous move, and shift the character at that index i times.
• Do nothing.

Shifting a character means replacing it by the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Shifting a character by i means applying the shift operations i times.

Remember that any index j can be picked at most once.

Return true if it's possible to convert s into t in no more than k moves, otherwise return false.

Example 1:

Input: s = "input", t = "ouput", k = 9
Output: true
Explanation: In the 6th move, we shift 'i' 6 times to get 'o'. And in the 7th move we shift 'n' to get 'u'.


Example 2:

Input: s = "abc", t = "bcd", k = 10
Output: false
Explanation: We need to shift each character in s one time to convert it into t. We can shift 'a' to 'b' during the 1st move. However, there is no way to shift the other characters in the remaining moves to obtain t from s.


Example 3:

Input: s = "aab", t = "bbb", k = 27
Output: true
Explanation: In the 1st move, we shift the first 'a' 1 time to get 'b'. In the 27th move, we shift the second 'a' 27 times to get 'b'.


Constraints:

• 1 <= s.length, t.length <= 10^5
• 0 <= k <= 10^9
• s, t contain only lowercase English letters.

Related Topics:
String, Greedy

## Solution 1.

• class Solution {
public boolean canConvertString(String s, String t, int k) {
if (s == null || t == null || s.length() != t.length())
return false;
int length = s.length();
int[] differences = new int[length];
for (int i = 0; i < length; i++) {
int difference = t.charAt(i) - s.charAt(i);
if (difference < 0)
difference += 26;
differences[i] = difference;
}
Arrays.sort(differences);
int index = 0;
while (index < length && differences[index] == 0)
index++;
if (index == length)
return true;
int prev = differences[index];
index++;
while (index < length) {
if (differences[index] == prev)
differences[index] = differences[index - 1] + 26;
else
prev = differences[index];
if (differences[index] > k)
return false;
index++;
}
return true;
}
}

############

class Solution {
public boolean canConvertString(String s, String t, int k) {
if (s.length() != t.length()) {
return false;
}
int[] cnt = new int[26];
for (int i = 0; i < s.length(); ++i) {
int x = (t.charAt(i) - s.charAt(i) + 26) % 26;
++cnt[x];
}
for (int i = 1; i < 26; ++i) {
if (i + 26 * (cnt[i] - 1) > k) {
return false;
}
}
return true;
}
}

• // OJ: https://leetcode.com/problems/can-convert-string-in-k-moves/
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool canConvertString(string s, string t, int k) {
if (s.size() != t.size()) return false;
int cnt[26] = {};
for (int i = 0; i < s.size(); ++i) {
int d = t[i] - s[i];
if (d == 0) continue;
if (d < 0) d += 26;
if (d + cnt[d]++ * 26 > k) return false;
}
return true;
}
};

• class Solution:
def canConvertString(self, s: str, t: str, k: int) -> bool:
if len(s) != len(t):
return False
cnt = [0] * 26
for a, b in zip(s, t):
x = (ord(b) - ord(a) + 26) % 26
cnt[x] += 1
for i in range(1, 26):
if i + 26 * (cnt[i] - 1) > k:
return False
return True


• func canConvertString(s string, t string, k int) bool {
if len(s) != len(t) {
return false
}
cnt := [26]int{}
for i := range s {
x := (t[i] - s[i] + 26) % 26
cnt[x]++
}
for i := 1; i < 26; i++ {
if i+26*(cnt[i]-1) > k {
return false
}
}
return true
}