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967. Numbers With Same Consecutive Differences

2022-09-05
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class Solution {
    public int[] numsSameConsecDiff(int n, int k) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 1; i < 10; i++) {
            backtrack(ans, i, 2, i + k, n, k);
            if (k != 0) {
                backtrack(ans, i, 2, i - k, n, k);
            }
        }
        int[] res = new int[ans.size()];
        for (int i = 0; i < ans.size(); i++) {
            res[i] = ans.get(i);
        }
        return res;
    }

    private void backtrack(List<Integer> ans, int num, int pos, int currDigit, int n, int k) {
        if (currDigit < 0 || currDigit > 9) {
            return;
        }
        int currNum = num * 10 + currDigit;
        if (pos == n) {
            ans.add(currNum);
            return;
        }
        backtrack(ans, currNum, pos + 1, currDigit + k, n, k);
        if (k != 0) {
            backtrack(ans, currNum, pos + 1, currDigit - k, n, k);
        }
    }
}

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912. Sort an Array

2022-09-05
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class Solution {
    public int[] sortArray(int[] nums) {
        mergeSort(nums, nums.clone(), 0, nums.length - 1);
        return nums;
    }

    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }

    // Bubble Sort 最大值到最后一个位置, 次大到倒数第二个...
    public void bubbleSort(int[] nums) {
        int inPlaceCount = 0;
        // We have nums.length numbers to process in order to let each of them be in
        // the right place
        for (int i = 0; i < nums.length - 1; i++) {
            boolean isSorted = true;
            // We don't bother the nums that are in place which is the last
            // "inPlaceCount" nums.
            for (int j = 0; j < nums.length - inPlaceCount - 1; j++) {
                if (nums[j] > nums[j + 1]) {
                    isSorted = false;
                    swap(nums, j, j + 1);
                }
            }
            // End the outer for loop earlier.
            if (isSorted)
                break;
            inPlaceCount += 1;
        }
        return nums;
    }

    // Selection Sort 第0个位置最小的坐,第1个位置次小的坐...
    public void selectionSort(int[] nums) {
        for (int i = 0; i < nums.length - 1; i++) {
            int minIdx = i;
            for (int j = i; j < nums.length; j++) {
                if (nums[j] < nums[minIdx]) {
                    minIdx = j;
                }
            }
            swap(nums, minIdx, i);
        }
        return nums;
    }

    // Insertion Sort 当前位置的值比前一个小,那就换到前面,一直到换不动为止
    public void insertionSort(int[] nums) {
        for (int i = 1; i < nums.length; i++) {
            for (int j = i; j > 0; j--) {
                if (nums[j] < nums[j - 1]) {
                    swap(nums, j, j - 1);
                } else {
                    break;
                }
            }
        }
        return nums;
    }

    // Quick Sort
    private void quickSort(int[] nums, int start, int end) {
        if (start >= end) {
            return;
        }
        int pivot = start, left = start + 1, right = end;
        while (left <= right) {
            if (nums[left] > nums[pivot] && nums[right] < nums[pivot]) {
                swap(nums, left, right);
                left += 1;
                right -= 1;
                continue;
            }
            if (nums[left] <= nums[pivot]) {
                left += 1;
            }
            if (nums[right] >= nums[pivot]) {
                right -= 1;
            }
        }
        swap(nums, pivot, right);
        boolean isLeftShorter = right - start <= end - right ? true : false;
        if (isLeftShorter) {
            quickSort(nums, start, right - 1);
            quickSort(nums, right + 1, end);
        } else {
            quickSort(nums, right + 1, end);
            quickSort(nums, start, right - 1);
        }
    }

    // Merge Sort 给一个范围, 能把这个范围内的数字给sort好. 注意nums和aux往下传递的时候要颠倒位置.
    private void mergeSort(int[] nums, int[] aux, int start, int end) {
        if (start >= end) {
            return;
        }
        int mid = start + (end - start) / 2;
        mergeSort(aux, nums, start, mid);
        mergeSort(aux, nums, mid + 1, end);
        merge(nums, aux, start, mid, end);
    }

    private void merge(int[] nums, int[] aux, int left, int middle, int right) {
        int ptr = left, ptrOne = left, ptrTwo = middle + 1;
        while (ptrOne <= middle && ptrTwo <= right) {
            if (aux[ptrOne] <= aux[ptrTwo]) {
                nums[ptr++] = aux[ptrOne++];
            } else {
                nums[ptr++] = aux[ptrTwo++];
            }
        }
        while (ptrOne <= middle) {
            nums[ptr++] = aux[ptrOne++];
        }
        while (ptrTwo <= right) {
            nums[ptr++] = aux[ptrTwo++];
        }
    }
}

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737. Sentence Similarity II

2022-09-05
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class Solution {
    public boolean areSentencesSimilarTwo(String[] sentence1, String[] sentence2, List<List<String>> similarPairs) {
        if (sentence1.length != sentence2.length) {
            return false;
        }
        Map<String, Set<String>> wordsSimilarSet = new HashMap<>();
        for (List<String> pair : similarPairs) {
            wordsSimilarSet.putIfAbsent(pair.get(0), new HashSet<>());
            wordsSimilarSet.get(pair.get(0)).add(pair.get(1));
            wordsSimilarSet.putIfAbsent(pair.get(1), new HashSet<>());
            wordsSimilarSet.get(pair.get(1)).add(pair.get(0));
        }
        for (int i = 0; i < sentence1.length; i++) {
            if (sentence1[i].equals(sentence2[i])) {
                continue;
            }
            if (!wordsSimilarSet.containsKey(sentence1[i]) || !wordsSimilarSet.containsKey(sentence2[i])) {
                return false;
            }
            if (!dfs(wordsSimilarSet, sentence1[i], sentence2[i], new HashSet<>())) {
                return false;
            }
        }
        return true;
    }

    private boolean dfs(Map<String, Set<String>> wordsSimilarSet, String word1, String word2, Set<String> visited) {
        if (wordsSimilarSet.get(word1).contains(word2)) {
            return true;
        }
        for (String child : wordsSimilarSet.get(word1)) {
            if (visited.add(child) && dfs(wordsSimilarSet, child, word2, visited)) {
                return true;
            }
        }
        return false;
    }
}

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547. Number of Provinces

2022-09-05
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class Solution {
    private int[] cityParent;
    private int[] rank;

    private void initialize() {
        for (int i = 1; i < cityParent.length; i++) {
            cityParent[i] = i;
            rank[i] = 1;
        }
    }

    private int find(int city) {
        if (city == cityParent[city]) {
            return city;
        }
        return cityParent[city] = find(cityParent[city]);
    }

    private void union(int city1, int city2) {
        int city1Root = find(city1);
        int city2Root = find(city2);
        if (city1Root != city2Root) {
            if (rank[city1Root] < rank[city2Root]) {
                cityParent[city1Root] = city2Root;
            } else if (rank[city1Root] > rank[city2Root]) {
                cityParent[city2Root] = city1Root;
            } else {
                cityParent[city2Root] = city1Root;
                rank[city1Root] += 1;
            }
        }
    }

    public int findCircleNum(int[][] isConnected) {
        int numOfCities = isConnected.length;
        cityParent = new int[numOfCities + 1];
        rank = new int[numOfCities + 1];
        initialize();
        for (int i = 0; i < isConnected.length; i++) {
            for (int j = 0; j < isConnected[0].length; j++) {
                if (isConnected[i][j] == 1) {
                    union(i + 1, j + 1);
                }
            }
        }
        Set<Integer> rootCities = new HashSet<>();
        for (int i = 1; i <= numOfCities; i++) {
            rootCities.add(find(i));
        }
        return rootCities.size();
    }
}

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453. Minimum Moves to Equal Array Elements

2022-09-05
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class Solution {
    public int minMoves(int[] nums) {
        int ans = 0;
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < nums.length; i++) {
            min = Math.min(nums[i], min);
        }
        for (int i = 0; i < nums.length; i++) {
            ans += nums[i] - min;
        }
        return ans;
    }
}

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429. N-ary Tree Level Order Traversal

2022-09-05
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class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> ans = new ArrayList<>();
        helper(root, ans, 0);
        return ans;
    }

    private void helper(Node node, List<List<Integer>> ans, int level) {
        if (node == null) {
            return;
        }
        if (level == ans.size()) {
            ans.add(new ArrayList<>());
        }
        ans.get(level).add(node.val);
        for (Node child : node.children) {
            helper(child, ans, level + 1);
        }
    }
}

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317. Shortest Distance from All Buildings

2022-09-05
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class Solution {
    private int[][] DIRECTIONS = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };

    public int shortestDistance(int[][] grid) {
        int[][][] gridSumCount = new int[grid.length][grid[0].length][2];
        Deque<int[]> buildings = new ArrayDeque<>();
        int buildingCount = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    buildings.offer(new int[] { i, j });
                    buildingCount += 1;
                }
            }
        }
        while (!buildings.isEmpty()) {
            int[] currBuilding = buildings.poll();
            Deque<int[]> possibleSites = new ArrayDeque<>();
            possibleSites.offer(currBuilding);
            boolean[][] visited = new boolean[grid.length][grid[0].length];
            visited[currBuilding[0]][currBuilding[1]] = true;
            int steps = 1;
            while (!possibleSites.isEmpty()) {
                int size = possibleSites.size();
                for (int i = 0; i < size; i++) {
                    int[] currPos = possibleSites.poll();
                    for (int[] direction : DIRECTIONS) {
                        int newRow = currPos[0] + direction[0];
                        int newCol = currPos[1] + direction[1];
                        if (!isOutOfBound(grid, newRow, newCol) && grid[newRow][newCol] == 0
                                && !visited[newRow][newCol]) {
                            gridSumCount[newRow][newCol][0] += steps;
                            gridSumCount[newRow][newCol][1] += 1;
                            visited[newRow][newCol] = true;
                            possibleSites.offer(new int[] { newRow, newCol });
                        }
                    }
                }
                steps += 1;
            }
        }
        int ans = Integer.MAX_VALUE;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 0 && gridSumCount[i][j][1] == buildingCount) {
                    ans = Math.min(ans, gridSumCount[i][j][0]);
                }
            }
        }
        return ans == Integer.MAX_VALUE ? -1 : ans;
    }

    private boolean isOutOfBound(int[][] grid, int row, int col) {
        return row < 0 || row >= grid.length || col < 0 || col >= grid[0].length;
    }
}

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187. Repeated DNA Sequences

2022-09-05
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class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        if (s.length() <= 10) {
            return new ArrayList<>();
        }
        int left = 0, right = 0, hash = 0;
        // Initialize the winodw
        for (right = 0; right < 10; right++) {
            int currDigit = getDigit(s.charAt(right));
            hash = hash * 10 + currDigit;
        }
        Map<Integer, Integer> visited = new HashMap<>();
        visited.put(hash, 1);
        List<String> ans = new ArrayList<>();
        int msbOffset = 1000000000;
        while (right < s.length()) {
            hash = (hash - msbOffset * (getDigit(s.charAt(left)))) * 10 + getDigit(s.charAt(right));
            if (visited.containsKey(hash) && visited.get(hash) == 1) {
                ans.add(s.substring(left + 1, right + 1));
            }
            visited.put(hash, visited.getOrDefault(hash, 0) + 1);
            left += 1;
            right += 1;
        }
        return ans;
    }

    private int getDigit(char c) {
        int currDigit = 0;
        switch (c) {
            case 'A':
                currDigit = 1;
                break;
            case 'C':
                currDigit = 2;
                break;
            case 'G':
                currDigit = 3;
                break;
            case 'T':
                currDigit = 4;
                break;
            default:
                currDigit = 0;
                break;
        }
        return currDigit;
    }
}

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Union Find

2022-09-03

Union Find Template (union by rank + path compression). Both techniques are used to optimize find method.
The below implementation is from LC disjoint set tutorial.

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875. Koko Eating Bananas

2022-09-03
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class Solution {
    public int minEatingSpeed(int[] piles, int h) {
        int left = 1, right = piles[0];
        for (int i = 1; i < piles.length; i++) {
            right = Math.max(right, piles[i]);
        }
        int ans = right;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            long count = 0;
            for (int pile : piles) {
                count += pile / mid;
                count += pile % mid == 0 ? 0 : 1;
                // count += Math.ceil((double) pile / mid);
            }
            if (count <= h) {
                ans = mid;
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
}

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