Some improvements and solution update

gouthampradhan · Jul 13, 2019 · 2627da3 · 2627da3
1 parent 6d17f4c
commit 2627da3
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diff --git a/.gitignore b/.gitignore
@@ -13,6 +13,5 @@ leetcode.ipr
 leetcode.iws
 out/
 problems/src.iml
-.idea/codeStyles/
-.idea/copyright/
-.idea/markdown-navigator/
+.idea/*
+.gradle/
diff --git a/problems/src/array/MaxConsecutiveOnes.java b/problems/src/array/MaxConsecutiveOnes.java
@@ -15,23 +15,22 @@ public static void main(String[] args) {
     //
   }
 
-    public int findMaxConsecutiveOnes(int[] nums) {
-        int max = 0;
-        boolean flag = false;
-        int count = 0;
-        for(int i = 0; i < nums.length; i ++){
-            if(nums[i] == 1){
-                if(!flag){
-                    flag = true;
-                }
-                count++;
-                max = Math.max(max, count);
-            } else{
-                count = 0;
-                flag = false;
-            }
+  public int findMaxConsecutiveOnes(int[] nums) {
+    int max = 0;
+    boolean flag = false;
+    int count = 0;
+    for (int i = 0; i < nums.length; i++) {
+      if (nums[i] == 1) {
+        if (!flag) {
+          flag = true;
         }
-        return max;
+        count++;
+        max = Math.max(max, count);
+      } else {
+        count = 0;
+        flag = false;
+      }
     }
-
+    return max;
+  }
 }
diff --git a/problems/src/array/MaxConsecutiveOnesII.java b/problems/src/array/MaxConsecutiveOnesII.java
@@ -13,60 +13,59 @@
  * stream? In other words, you can't store all numbers coming from the stream as it's too large to
  * hold in memory. Could you solve it efficiently?
  *
- * Solution: O(N)
- * Maintain a left and right auxiliary array with counts of contagious 1's from both directions.
- * Now, iterate through the array and flip a 0 to 1 and sum up left and right contagious sum of 1's and return the
- * max sum as the answer
+ * <p>Solution: O(N) Maintain a left and right auxiliary array with counts of contagious 1's from
+ * both directions. Now, iterate through the array and flip a 0 to 1 and sum up left and right
+ * contagious sum of 1's and return the max sum as the answer
  */
 public class MaxConsecutiveOnesII {
   public static void main(String[] args) {
     //
   }
-    public int findMaxConsecutiveOnes(int[] nums) {
-        int[] L = new int[nums.length];
-        int[] R = new int[nums.length];
-        boolean flag = false;
-        int count = 0;
-        int max = 0;
-        for(int j = 0; j < nums.length; j ++){
-            if(nums[j] == 1){
-                if(!flag){
-                    flag = true;
-                }
-                count++;
-                L[j] = count;
-            } else{
-                count = 0;
-                flag = false;
-                L[j] = count;
-            }
-            max = Math.max(max, count);
-        }
 
-        flag = false;
-        count = 0;
-        for(int j = nums.length - 1; j >= 0; j --){
-            if(nums[j] == 1){
-                if(!flag){
-                    flag = true;
-                }
-                count++;
-                R[j] = count;
-            } else{
-                count = 0;
-                flag = false;
-                R[j] = count;
-            }
+  public int findMaxConsecutiveOnes(int[] nums) {
+    int[] L = new int[nums.length];
+    int[] R = new int[nums.length];
+    boolean flag = false;
+    int count = 0;
+    int max = 0;
+    for (int j = 0; j < nums.length; j++) {
+      if (nums[j] == 1) {
+        if (!flag) {
+          flag = true;
         }
+        count++;
+        L[j] = count;
+      } else {
+        count = 0;
+        flag = false;
+        L[j] = count;
+      }
+      max = Math.max(max, count);
+    }
 
-        for(int i = 0; i < nums.length; i ++){
-            if(nums[i] == 0){
-                int l = i == 0 ? 0 : L[i - 1];
-                int r = i == nums.length - 1 ? 0 : R[i + 1];
-                max = Math.max(max, l + r + 1);
-            }
+    flag = false;
+    count = 0;
+    for (int j = nums.length - 1; j >= 0; j--) {
+      if (nums[j] == 1) {
+        if (!flag) {
+          flag = true;
         }
-        return max;
+        count++;
+        R[j] = count;
+      } else {
+        count = 0;
+        flag = false;
+        R[j] = count;
+      }
     }
 
+    for (int i = 0; i < nums.length; i++) {
+      if (nums[i] == 0) {
+        int l = i == 0 ? 0 : L[i - 1];
+        int r = i == nums.length - 1 ? 0 : R[i + 1];
+        max = Math.max(max, l + r + 1);
+      }
+    }
+    return max;
+  }
 }
diff --git a/problems/src/backtracking/MatchsticksToSquare.java b/problems/src/backtracking/MatchsticksToSquare.java
@@ -1,5 +1,7 @@
 package backtracking;
+
 import java.util.*;
+
 /**
  * Created by gouthamvidyapradhan on 25/05/2019 Remember the story of Little Match Girl? By now, you
  * know exactly what matchsticks the little match girl has, please find out a way you can make one
@@ -19,94 +21,96 @@
  * sum of the given matchsticks is in the range of 0 to 10^9. The length of the given matchstick
  * array will not exceed 15.
  *
- * Solution: O(2 ^ N): Generate a power set of all combination of numbers for the given array which sum up to the
- * length of a side of square.
- * Now, to check if a square can be made using all the sides sticks of different length, generate a hash for for each of
- * the combination which was generated in the previous step. The hash function should be such that it uses unique
- * indexes of each match stick. If 4 different hash values are formed using unique and all indices then a square is
- * possible.
+ * <p>Solution: O(2 ^ N): Generate a power set of all combination of numbers for the given array
+ * which sum up to the length of a side of square. Now, to check if a square can be made using all
+ * the sides sticks of different length, generate a hash for for each of the combination which was
+ * generated in the previous step. The hash function should be such that it uses unique indexes of
+ * each match stick. If 4 different hash values are formed using unique and all indices then a
+ * square is possible.
  */
 public class MatchsticksToSquare {
-    /**
-     * Main method
-     * @param args
-     */
+  /**
+   * Main method
+   *
+   * @param args
+   */
   public static void main(String[] args) {
-      int[] A = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 6, 10, 10};
+    int[] A = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 6, 10, 10};
     System.out.println(new MatchsticksToSquare().makesquare(A));
   }
 
-    class Pair {
-        int value, i;
-        Pair(int value, int i){
-            this.value = value;
-            this.i = i;
-        }
+  class Pair {
+    int value, i;
+
+    Pair(int value, int i) {
+      this.value = value;
+      this.i = i;
     }
+  }
 
-    public boolean makesquare(int[] nums) {
-      if(nums.length == 0) return false;
-        int sum = 0;
-        for(int n : nums){
-            sum += n;
-        }
-        int side = sum / 4;
-        if((sum % 4) != 0) return false;
-        List<List<Pair>> list = powerSet(nums, side);
-        Set<Integer> hashIndex = new HashSet<>();
-        int cons = 0;
-        for(int i = 0; i < nums.length; i ++){
-            cons |= (1 << i);
+  public boolean makesquare(int[] nums) {
+    if (nums.length == 0) return false;
+    int sum = 0;
+    for (int n : nums) {
+      sum += n;
+    }
+    int side = sum / 4;
+    if ((sum % 4) != 0) return false;
+    List<List<Pair>> list = powerSet(nums, side);
+    Set<Integer> hashIndex = new HashSet<>();
+    int cons = 0;
+    for (int i = 0; i < nums.length; i++) {
+      cons |= (1 << i);
+    }
+    for (int i = 0; i < list.size(); i++) {
+      for (int j = i + 1; j < list.size(); j++) {
+        Set<Integer> indexList = new HashSet<>();
+        List<Pair> list1 = list.get(i);
+        List<Pair> list2 = list.get(j);
+        int hash = 0;
+        for (Pair l1 : list1) {
+          indexList.add(l1.i);
+          hash |= (1 << l1.i);
         }
-        for(int i = 0; i < list.size(); i ++){
-          for(int j = i + 1; j < list.size(); j ++){
-            Set<Integer> indexList = new HashSet<>();
-            List<Pair> list1 =  list.get(i);
-            List<Pair> list2 =  list.get(j);
-            int hash = 0;
-            for(Pair l1 : list1){
-                indexList.add(l1.i);
-                hash |= (1 << l1.i);
-            }
-            boolean allUnique = true;
-            for(Pair l2 : list2){
-                if(indexList.contains(l2.i)) {
-                    allUnique = false;
-                    break;
-                }
-                indexList.add(l2.i);
-                hash |= (1 << l2.i);
-            }
-            if(allUnique){
-                hashIndex.add(hash);
-                int complement = ((~ hash) & cons);
-                if(hashIndex.contains(complement)) return true;
-            }
+        boolean allUnique = true;
+        for (Pair l2 : list2) {
+          if (indexList.contains(l2.i)) {
+            allUnique = false;
+            break;
           }
+          indexList.add(l2.i);
+          hash |= (1 << l2.i);
+        }
+        if (allUnique) {
+          hashIndex.add(hash);
+          int complement = ((~hash) & cons);
+          if (hashIndex.contains(complement)) return true;
+        }
       }
-      return false;
     }
+    return false;
+  }
 
-    private List<List<Pair>> powerSet(int[] nums, int expectedSum){
-      List<List<Pair>> result = new ArrayList<>();
-      generate(0, nums, new ArrayList<>(), result, 0, expectedSum);
-      return result;
-    }
+  private List<List<Pair>> powerSet(int[] nums, int expectedSum) {
+    List<List<Pair>> result = new ArrayList<>();
+    generate(0, nums, new ArrayList<>(), result, 0, expectedSum);
+    return result;
+  }
 
-    private void generate(int i, int[] nums, List<Pair> subList, List<List<Pair>> result, int sum,
-                          int expected){
-      if(i >= nums.length){
-          if(sum == expected){
-              List<Pair> pairs = new ArrayList<>(subList);
-              result.add(pairs);
-          }
-      } else{
-          if(sum + nums[i] <= expected){
-              subList.add(new Pair(nums[i], i));
-              generate(i + 1, nums, subList, result, sum + nums[i], expected);
-              subList.remove(subList.size() - 1);
-          }
-          generate(i + 1, nums, subList, result, sum, expected);
+  private void generate(
+      int i, int[] nums, List<Pair> subList, List<List<Pair>> result, int sum, int expected) {
+    if (i >= nums.length) {
+      if (sum == expected) {
+        List<Pair> pairs = new ArrayList<>(subList);
+        result.add(pairs);
       }
+    } else {
+      if (sum + nums[i] <= expected) {
+        subList.add(new Pair(nums[i], i));
+        generate(i + 1, nums, subList, result, sum + nums[i], expected);
+        subList.remove(subList.size() - 1);
+      }
+      generate(i + 1, nums, subList, result, sum, expected);
     }
+  }
 }