[GangBean] Week 3

combination-sum/GangBean.java

@@ -0,0 +1,43 @@
+class Solution {
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        /**
+        1. understanding
+        - find all combinations, which sum to target
+        - can use same number multiple times
+        2. strategy
+        - dp[target]: all combination, which sum to target
+        - dp[n + 1] = dp[n] | dp[1]
+        - [2,3,6,7], target = 7
+            - dp[0] = [[]]
+            - dp[1] = [[]]
+            - dp[2] = [[2]]
+            - dp[3] = [[3]]
+            - dp[4] = dp[2] | dp[2] = [[2,2]]
+            - dp[5] = dp[2] | dp[3] = [[2,3]]
+            - dp[6] = dp[2] | dp[4] , dp[3] | dp[3] = [[2,2,2], [3,3]]
+            - dp[7] = dp[2] | dp[5], dp[3] | dp[4], dp[6] | dp[1], dp[7] = [[2,2,3],]
+        3. complexity
+        - time: O(target * N) where N is length of candidates
+        - space: O(target * N)
+        */
+        List<List<Integer>>[] dp = new List[target + 1];
+        for (int i = 0; i <= target; i++) {
+            dp[i] = new ArrayList<>();
+        }
+
+        dp[0].add(new ArrayList<>());
+
+        for (int candidate : candidates) {
+            for (int i = candidate; i <= target; i++) {
+                for (List<Integer> combination : dp[i - candidate]) {
+                    List<Integer> newCombination = new ArrayList<>(combination);
+                    newCombination.add(candidate);
+                    dp[i].add(newCombination);
+                }
+            }
+        }
+
+        return dp[target];
+    }
+}
+

maximum-subarray/GangBean.java

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+class Solution {
+    public int maxSubArray(int[] nums) {
+        /**
+        1. understanding
+        - integer array nums
+        - find largest subarray sum
+        2. starategy
+        - calculate cumulative sum
+        - mem[i+1] = num[i+1] + mem[i] if (num[i+1] + mem[i] >= 0) else num[i+1]
+        3. complexity
+        - time: O(N)
+        - space: O(1)
+        */
+        int prev = 0;
+        int curr = 0;
+        int max = Integer.MIN_VALUE;
+        for (int i = 0 ; i < nums.length; i++) {
+            curr = nums[i];
+            if (prev >= 0) {
+                curr += prev;
+            }
+            max = Math.max(max, curr);
+            prev = curr;
+        }
+        return max;
+    }
+}
+

product-of-array-except-self/GangBean.java

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+class Solution {
+    public int[] productExceptSelf(int[] nums) {
+        /**
+            1. understanding
+            - given integer array nums
+            - product of which's all elements except that element
+            - should be under O(N) time complexity
+            - should not use division operation
+            2. strategy
+            - brute force: O(n^2)
+                - for every elements, calculate product of other elements
+                    - 1: 2 * [3 * 4]
+                    - 2: 1 * [3 * 4]
+                    - 3: [1 * 2] * 4
+                    - 4: [1 * 2] * 3
+            - dynamic programming
+                - assign array mem to memorize product till that idx
+                - mem memorize in ascending order product value and reverse order product value
+            3. complexity
+            - time: O(N)
+            - space: O(N)
+        */
+        // 1. assign array variable mem
+        int[][] mem = new int[nums.length][];
+        for (int i = 0 ; i < nums.length; i++) {
+            mem[i] = new int[2];
+        }
+
+        // 2. calculate product values
+        for (int i = 0 ; i < nums.length; i++) { // O(N)
+            if (i == 0) {
+                mem[i][0] = nums[i];
+                continue;
+            }
+            mem[i][0] = nums[i] * mem[i-1][0];
+        }
+
+        for (int i = nums.length - 1; i >= 0; i--) { // O(N)
+            if (i == nums.length - 1) {
+                mem[i][1] = nums[i];
+                continue;
+            }
+            mem[i][1] = nums[i] * mem[i+1][1];
+        }
+
+        int[] ret = new int[nums.length];
+        for (int i = 0; i < nums.length; i++) { // O(N)
+            int left = (i - 1) >= 0 ? mem[i-1][0] : 1;
+            int right = (i + 1) < nums.length ? mem[i+1][1] : 1;
+            ret[i] = left * right;
+        }
+
+        return ret;
+    }
+}
+

reverse-bits/GangBean.java

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+public class Solution {
+    // you need treat n as an unsigned value
+    public int reverseBits(int n) {
+        /**
+        1. understanding
+        - return the reversed 32 bit input num
+        2. strategy
+        - assign stack
+        - iterate until stack.size is 32
+        - push value % 2, and value /= 2
+        3. complexity
+        - time: O(1)
+        - space: O(1)
+        */
+        int reversed = 0;
+        for (int i = 0; i < 32; i++) {
+            reversed <<= 1;
+            reversed |= n & 1;
+            n >>= 1;
+        }
+        return reversed;
+    }
+}
+

two-sum/GangBean.java

@@ -0,0 +1,35 @@
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        /**
+            1. understanding
+            - find the indices where the numbers of that index pair sum upto the target number.
+            - exactly one solution
+            - use each element only once
+            - return in any order
+            2. strategy
+            - brute force:
+                - validate for all pair sum
+            - hashtable:
+                - assing hashtable variable to save the nums and index
+                - while iterate over the nums,
+                - calculate the diff, and check if the diff in the hashtable.
+                - or save new entry.
+            3. complexity
+            - time: O(N)
+            - space: O(N)
+        */
+
+        Map<Integer, Integer> map = new HashMap<>();
+
+        for (int i = 0; i < nums.length; i++) {
+            int diff = target - nums[i];
+            if (map.containsKey(diff)) {
+                return new int[] {map.get(diff), i};
+            }
+            map.put(nums[i], i);
+        }
+
+        return new int[] {};
+    }
+}
+