shifting letters

sherxon · Jul 28, 2018 · 03a5897 · 03a5897
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diff --git a/src/problems/Medium.txt b/src/problems/Medium.txt
@@ -2800,4 +2800,28 @@ Explanation: Alice's hand can't be rearranged into groups of 4.
 
  1 <= hand.length <= 10000
  0 <= hand[i] <= 10^9
- 1 <= W <= hand.length
+ 1 <= W <= hand.length
+ 190)Shifting Letters
+ We have a string S of lowercase letters, and an integer array shifts.
+
+ Call the shift of a letter, the next letter in the alphabet, (wrapping around so that 'z' becomes 'a').
+
+ For example, shift('a') = 'b', shift('t') = 'u', and shift('z') = 'a'.
+
+ Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times.
+
+ Return the final string after all such shifts to S are applied.
+
+ Example 1:
+
+ Input: S = "abc", shifts = [3,5,9]
+ Output: "rpl"
+ Explanation:
+ We start with "abc".
+ After shifting the first 1 letters of S by 3, we have "dbc".
+ After shifting the first 2 letters of S by 5, we have "igc".
+ After shifting the first 3 letters of S by 9, we have "rpl", the answer.
+ Note:
+
+ 1 <= S.length = shifts.length <= 20000
+ 0 <= shifts[i] <= 10 ^ 9
diff --git a/src/problems/medium/ShiftingLetters.java b/src/problems/medium/ShiftingLetters.java
@@ -0,0 +1,22 @@
+package problems.medium;
+
+/**
+ * Why Did you create this class? what does it do?
+ */
+public class ShiftingLetters {
+    public String shiftingLetters(String s, int[] a) {
+        if (a == null || a.length == 0)
+            return s;
+        a[a.length - 1] %= 26;
+        for (int i = a.length - 2; i >= 0; i--) {
+            a[i] = (a[i] + a[i + 1]) % 26;
+        }
+        char[] c = s.toCharArray();
+        for (int i = 0; i < a.length; i++) {
+            c[i] = (char) (a[i] + c[i]);
+            if (c[i] > 'z')
+                c[i] = (char) (c[i] % 'z' + 'a' - 1);
+        }
+        return new String(c);
+    }
+}