By Long Luo

This article is the solution Priority Queue + Greedy: Consider Ladder as a One-time Unlimited Bricks of Problem 1642. Furthest Building You Can Reach.

Intuition

While we are moving, we will need to use bricks or ladders several times. Suppose we have to move to the next building, which means we must use one ladder or \(\Delta h\) bricks, so there is a question, whether to use ladders or bricks?

If the gap of buildings is large, we may use the ladder, otherwise we use the bricks.

We can consider ladder as a one-time unlimited number of bricks, That is, if we have l ladders, we will use the ladder on the \(l\) times where \(\Delta h\) is the largest, and use the bricks in the rest.

Therefore, we got the answer. We maintain no more than \(l\) largest \(\Delta h\) using priority queues, and these are where the ladders are used. For the remaining \(\Delta h\), we need to use bricks, so we need to accumulate them, if the \(sum\) exceeds the number of bricks \(b\), then we have move to the furthest building.

The code is as follows:

By Long Luo

This article is the solution 2 Approaches: DFS and BFS with Detailed Explanation of Problem 199. Binary Tree Right Side View.

Here shows 2 Approaches to slove this problem: DFS and BFS.

DFS

We traverse the tree in the order of \(\textit{root node} \to \textit{right subtree} \to \textit{left subtree}\) to ensure that each level traverse the rightmost node first.

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public List<Integer> rightSideView_dfs(TreeNode root) {
    List<Integer> ans = new ArrayList<>();
    if (root == null) {
        return ans;
    }

    dfs(root, 1, ans);
    return ans;
}

public void dfs(TreeNode root, int depth, List<Integer> numList) {
    if (root == null) {
        return;
    }

    if (numList.size() < depth) {
        numList.add(root.val);
    }

    dfs(root.right, depth + 1, numList);
    dfs(root.left, depth + 1, numList);
}

Analysis

• Time Complexity: \(O(n)\)
• Space Complexity: \(O(n)\)

BFS

We can use BFS to traverse the levels of the tree and record the last element of each level.

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public List<Integer> rightSideView_bfs(TreeNode root) {
    List<Integer> ans = new ArrayList<>();
    if (root == null) {
        return ans;
    }

    Queue<TreeNode> queue = new LinkedList<>();
    queue.offer(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        for (int i = size - 1; i >= 0; i--) {
            TreeNode node = queue.poll();
            if (i == size - 1) {
                ans.add(node.val);
            }

            if (node.right != null) {
                queue.offer(node.right);
            }

            if (node.left != null) {
                queue.offer(node.left);
            }
        }
    }

    return ans;
}

Analysis

• Time Complexity: \(O(n)\).
• Space Complexity: \(O(n)\).

All suggestions are welcome. If you have any query or suggestion please comment below. Please upvote👍 if you like💗 it. Thank you:-)

Explore More Leetcode Solutions. 😉😃💗

By Long Luo

This article is the solution 3 Approaches: DFS, BFS, DP of Problem 322. Coin Change.

Here shows 3 Approaches to slove this problem: DFS, BFS and Dynamic Programming.

DFS

TLE!

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class Solution {
    int minCount = Integer.MAX_VALUE;

    public int coinChange(int[] coins, int amount) {
        if (amount <= 0) {
            return 0;
        }

        minCount = Integer.MAX_VALUE;
        dfs(coins, amount, 0);
        return minCount == Integer.MAX_VALUE ? -1 : minCount;
    }

    private int dfs(int[] coins, int remain, int count) {
        if (remain < 0) {
            return -1;
        }

        if (remain == 0) {
            minCount = Math.min(minCount, count);
            return count;
        }
        
        for (int x : coins) {
            dfs(coins, remain - x, count + 1);
        }

        return -1;
    }
}

Sorting the array first, then use DFS:

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class Solution {
    int minAns = Integer.MAX_VALUE;
    
    public int coinChange(int[] coins, int amount) {
        Arrays.sort(coins);
        minAns = Integer.MAX_VALUE;
        coinChange(coins, amount, coins.length - 1, 0);
        return minAns == Integer.MAX_VALUE ? -1 : minAns;
    }

    private void coinChange(int[] coins, int amount, int index, int cnt) {
        if (amount == 0) {
            minAns = Math.min(minAns, cnt);
            return;
        }

        if (index < 0) {
            return;
        }

        for (int k = amount / coins[index]; k >= 0 && k + cnt < minAns; k--) {
            coinChange(coins, amount - k * coins[index], index - 1, cnt + k);
        }
    }
}

Memory DFS:

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public class Solution {
    public int coinChange(int[] coins, int amount) {
        if (amount <= 0) {
            return 0;
        }

        return coinChange(coins, amount, new int[amount]);
    }

    private int coinChange(int[] coins, int rem, int[] count) {
        if (rem < 0) {
            return -1;
        }

        if (rem == 0) {
            return 0;
        }

        if (count[rem - 1] != 0) {
            return count[rem - 1];
        }

        int min = Integer.MAX_VALUE;
        for (int coin : coins) {
            int res = coinChange(coins, rem - coin, count);
            if (res >= 0 && res < min) {
                min = 1 + res;
            }
        }
        count[rem - 1] = (min == Integer.MAX_VALUE) ? -1 : min;
        return count[rem - 1];
    }
}

Analysis

• Time Complexity: \(O(amount \times n)\)
• Space Complexity: \(O(amount)\)

By Long Luo

This article is the solution Dynamic Programming Space O(n) Solutions: Top-Down and Bottom-Up Approaches of Problem 120. Triangle .

Intuition

This problem is a classic and typical dynamic programming problem. We can break the large problem into sub-problems.

We can use both the Top-Down and Bottom-Up approach to solve this problem.

Top-Down Approach

1. State definition:

\(dp[i][j]\) represents the minimum path sum of row \(i\) and column \(j\).

2. State Analysis:

\(dp[0][0]=c[0][0]\)

3. The State Transfer Equation:

\[ dp[i][j] = \begin{cases} dp[i-1][0] + c[i][0] & j=0 \\ dp[i-1][i-1] + c[i][i] & j==i \\ min(dp[i-1][j-1], dp[i-1][j]) + c[i][j] & 0 < j < i \\ \end{cases} \]

so we can easily write such code:

By Long Luo

This article is the solution How to Maintain the Enqueued Tasks? | Sorting the array then Priority Queue | minHeap of Problem 1834. Single-Threaded CPU .

Intuition

To simulate the CPU operations, there comes \(3\) questions:

1. Which task should int the enqueued tasks list?

2. How to maintain the enqueued tasks?

3. Which task of the enqueued tasks should the CPU choose?

Let’s answer the \(3\) questions:

1. We assign the tasks to the CPU by \(\textit{enqueueTime}\), so we sort the array first by \(\textit{enqueueTime}\). However, we will lose the \(\textit{index}\) of the task.

We can parse the task by creating a new class \(\texttt{Job}\), whose members are \(\textit{id}\), \(\textit{enqueueTime}\), \(\textit{processTime}\).

2. We put all the tasks assigned to the CPU into a Priority Queue and poll out the task whose \(\textit{processTime}\) is the least each time.

3. We can maintain a \(\textit{curTime}\) variable, which represents the current time with initial value is \(0\).

If the CPU has no tasks to execute, we set the \(\textit{curTime}\) to \(\textit{enqueueTime}\) of the next task in the array that has not yet been assigned to the CPU.

After this, we put all \(\textit{enqueueTime} \le \textit{curTime}\) tasks into the Priority Queue.