Added 1 solution & refactored 2 solutions

varunu28 · varunu28 · commit 524d18f580a0 · 2019-06-08T17:38:36.000-07:00
diff --git a/Easy/Array Partition I.java b/Easy/Array Partition I.java
@@ -0,0 +1,11 @@
+class Solution {
+    public int arrayPairSum(int[] nums) {
+        Arrays.sort(nums);
+        int sum = 0;
+        for (int i = 0; i < nums.length; i += 2) {
+            sum += nums[i];
+        }
+        
+        return sum;
+    }
+}
diff --git a/Easy/Array Partition.java b/Easy/Array Partition.java
diff --git a/Easy/Fixed Point.java b/Easy/Fixed Point.java
@@ -0,0 +1,18 @@
+class Solution {
+    public int fixedPoint(int[] A) {
+        int low = 0;
+        int high = A.length - 1;
+        
+        while (low < high) {
+            int mid = (low + high) / 2;
+            if (A[mid] - mid < 0) {
+                low = mid + 1;
+            }
+            else {
+                high = mid;
+            }
+        }
+        
+        return low == A[low] ? low : -1;
+    }
+}
diff --git a/Medium/Custom Sort String.java b/Medium/Custom Sort String.java
@@ -1,47 +1,27 @@
 class Solution {
-    public String customSortString(String s, String t) {
-
-        // Separate out the characters from t which are not part of s
-        StringBuilder nonS = new StringBuilder();
-        for (int i=0; i<t.length(); i++) {
-            if (s.indexOf(t.charAt(i)) == -1) {
-                nonS.append(t.charAt(i));
-            }
+    public String customSortString(String S, String T) {
+        int[] counter = new int[26];
+        for (char c : T.toCharArray()) {
+            counter[c - 'a']++;
         }
-
-        // Create a character array of t with words present in s
-        Character[] arr = new Character[t.length() - nonS.length()];
-        for (int i=0,j=0; i<t.length(); i++) {
-            if (s.indexOf(t.charAt(i)) != -1) {
-                arr[j++] = t.charAt(i);
+        
+        StringBuilder sb = new StringBuilder();
+        for (char c : S.toCharArray()) {
+            int count = counter[c - 'a'];
+            for (int i = 0; i < count; i++) {
+                sb.append(c);
             }
+            
+            counter[c - 'a'] = 0;
         }
-
-        // Create a value map for characters in s with increasing counter
-        int counter = 0;
-        Map<Character, Integer> map = new HashMap<>();
-        for (char c : s.toCharArray()) {
-            map.put(c, counter);
-            counter++;
-        }
-
-        // Sort the character array based on value map
-        Arrays.sort(arr, new Comparator<Character>() {
-            @Override
-            public int compare(Character o1, Character o2) {
-                return map.get(o1) - map.get(o2);
+        
+        for (char c ='a'; c <= 'z'; c++) {
+            int count = counter[c - 'a'];
+            for (int i = 0; i < count; i++) {
+                sb.append(c);
             }
-        });
-
-        // Prepare the final string answer
-        StringBuilder sb = new StringBuilder();
-        for (Character c : arr) {
-            sb.append(c);
         }
-
-        // Append characters not present in s to the end of solution
-        sb.append(nonS.toString());
-
+        
         return sb.toString();
     }
 }
