3466. Maximum Coin Collection 🔒

Description

Mario drives on a two-lane freeway with coins every mile. You are given two integer arrays, lane1 and lane2, where the value at the i^th index represents the number of coins he gains or loses in the i^th mile in that lane.

If Mario is in lane 1 at mile i and lane1[i] > 0, Mario gains lane1[i] coins.
If Mario is in lane 1 at mile i and lane1[i] < 0, Mario pays a toll and loses abs(lane1[i]) coins.
The same rules apply for lane2.

Mario can enter the freeway anywhere and exit anytime after traveling at least one mile. Mario always enters the freeway on lane 1 but can switch lanes at most 2 times.

A lane switch is when Mario goes from lane 1 to lane 2 or vice versa.

Return the maximum number of coins Mario can earn after performing at most 2 lane switches.

Note: Mario can switch lanes immediately upon entering or just before exiting the freeway.

Example 1:

Input: lane1 = [1,-2,-10,3], lane2 = [-5,10,0,1]

Output: 14

Explanation:

Mario drives the first mile on lane 1.
He then changes to lane 2 and drives for two miles.
He changes back to lane 1 for the last mile.

Mario collects 1 + 10 + 0 + 3 = 14 coins.

Example 2:

Input: lane1 = [1,-1,-1,-1], lane2 = [0,3,4,-5]

Output: 8

Explanation:

Mario starts at mile 0 in lane 1 and drives one mile.
He then changes to lane 2 and drives for two more miles. He exits the freeway before mile 3.

He collects 1 + 3 + 4 = 8 coins.

Example 3:

Input: lane1 = [-5,-4,-3], lane2 = [-1,2,3]

Output: 5

Explanation:

Mario enters at mile 1 and immediately switches to lane 2. He stays here the entire way.

He collects a total of 2 + 3 = 5 coins.

Example 4:

Input: lane1 = [-3,-3,-3], lane2 = [9,-2,4]

Output: 11

Explanation:

Mario starts at the beginning of the freeway and immediately switches to lane 2. He stays here the whole way.

He collects a total of 9 + (-2) + 4 = 11 coins.

Example 5:

Input: lane1 = [-10], lane2 = [-2]

Output: -2

Explanation:

Since Mario must ride on the freeway for at least one mile, he rides just one mile in lane 2.

He collects a total of -2 coins.

Constraints:

1 <= lane1.length == lane2.length <= 10⁵
-10⁹ <= lane1[i], lane2[i] <= 10⁹

Solutions

Solution 1: Memoized Search

We design a function $dfs (i, j, k) <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext mathvariant="italic">dfs</mtext><mo stretchy="false">(</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo>,</mo><mi>k</mi><mo stretchy="false">)</mo></math>$ , which represents the maximum number of coins Mario can collect starting from position $i <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi></math>$ , currently on lane $j <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>j</mi></math>$ , with $k <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>k</mi></math>$ lane changes remaining. The answer is the maximum value of $dfs (i, 0, 2) <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext mathvariant="italic">dfs</mtext><mo stretchy="false">(</mo><mi>i</mi><mo>,</mo><mn>0</mn><mo>,</mo><mn>2</mn><mo stretchy="false">)</mo></math>$ for all $i <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi></math>$ .

The function $dfs (i, j, k) <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext mathvariant="italic">dfs</mtext><mo stretchy="false">(</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo>,</mo><mi>k</mi><mo stretchy="false">)</mo></math>$ is calculated as follows:

If $i \geq n <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>i</mi><mo>\geq</mo><mi>n</mi></math>$ , it means Mario has reached the end, return 0;
If no lane change is made, Mario can drive 1 mile, then exit, or continue driving, taking the maximum of the two, i.e., $max (x, dfs (i + 1, j, k) + x) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo data-mjx-texclass="OP" movablelimits="true">max</mo><mo stretchy="false">(</mo><mi>x</mi><mo>,</mo><mtext mathvariant="italic">dfs</mtext><mo stretchy="false">(</mo><mi>i</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>j</mi><mo>,</mo><mi>k</mi><mo stretchy="false">)</mo><mo>+</mo><mi>x</mi><mo stretchy="false">)</mo></math>$ ;
If a lane change is possible, there are two choices: drive 1 mile and then change lanes, or change lanes directly, taking the maximum of these two cases, i.e., $max (dfs (i + 1, j \oplus 1, k - 1) + x, dfs (i, j \oplus 1, k - 1)) <math xmlns="http://www.w3.org/1998/Math/MathML"><mo data-mjx-texclass="OP" movablelimits="true">max</mo><mo stretchy="false">(</mo><mtext mathvariant="italic">dfs</mtext><mo stretchy="false">(</mo><mi>i</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>j</mi><mo>\oplus</mo><mn>1</mn><mo>,</mo><mi>k</mi><mo>-</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><mi>x</mi><mo>,</mo><mtext mathvariant="italic">dfs</mtext><mo stretchy="false">(</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo>\oplus</mo><mn>1</mn><mo>,</mo><mi>k</mi><mo>-</mo><mn>1</mn><mo stretchy="false">)</mo><mo stretchy="false">)</mo></math>$ .
Where $x <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>x</mi></math>$ represents the number of coins at the current position.

To avoid repeated calculations, we use memoized search to store the results that have already been computed.

Time complexity is $O (n) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></math>$ , and space complexity is $O (n) <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></math>$ . Where $n <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>n</mi></math>$ represents the length of the lanes.

class Solution {
    private int n;
    private int[] lane1;
    private int[] lane2;
    private Long[][][] f;

    public long maxCoins(int[] lane1, int[] lane2) {
        n = lane1.length;
        this.lane1 = lane1;
        this.lane2 = lane2;
        f = new Long[n][2][3];
        long ans = Long.MIN_VALUE;
        for (int i = 0; i < n; ++i) {
            ans = Math.max(ans, dfs(i, 0, 2));
        }
        return ans;
    }

    private long dfs(int i, int j, int k) {
        if (i >= n) {
            return 0;
        }
        if (f[i][j][k] != null) {
            return f[i][j][k];
        }
        int x = j == 0 ? lane1[i] : lane2[i];
        long ans = Math.max(x, dfs(i + 1, j, k) + x);
        if (k > 0) {
            ans = Math.max(ans, dfs(i + 1, j ^ 1, k - 1) + x);
            ans = Math.max(ans, dfs(i, j ^ 1, k - 1));
        }
        return f[i][j][k] = ans;
    }
}

class Solution {
public:
    long long maxCoins(vector<int>& lane1, vector<int>& lane2) {
        int n = lane1.size();
        long long ans = -1e18;
        vector<vector<vector<long long>>> f(n, vector<vector<long long>>(2, vector<long long>(3, -1e18)));
        auto dfs = [&](this auto&& dfs, int i, int j, int k) -> long long {
            if (i >= n) {
                return 0LL;
            }
            if (f[i][j][k] != -1e18) {
                return f[i][j][k];
            }
            int x = j == 0 ? lane1[i] : lane2[i];
            long long ans = max((long long) x, dfs(i + 1, j, k) + x);
            if (k > 0) {
                ans = max(ans, dfs(i + 1, j ^ 1, k - 1) + x);
                ans = max(ans, dfs(i, j ^ 1, k - 1));
            }
            return f[i][j][k] = ans;
        };
        for (int i = 0; i < n; ++i) {
            ans = max(ans, dfs(i, 0, 2));
        }
        return ans;
    }
};

class Solution:
    def maxCoins(self, lane1: List[int], lane2: List[int]) -> int:
        @cache
        def dfs(i: int, j: int, k: int) -> int:
            if i >= n:
                return 0
            x = lane1[i] if j == 0 else lane2[i]
            ans = max(x, dfs(i + 1, j, k) + x)
            if k > 0:
                ans = max(ans, dfs(i + 1, j ^ 1, k - 1) + x)
                ans = max(ans, dfs(i, j ^ 1, k - 1))
            return ans

        n = len(lane1)
        ans = -inf
        for i in range(n):
            ans = max(ans, dfs(i, 0, 2))
        return ans

func maxCoins(lane1 []int, lane2 []int) int64 {
	n := len(lane1)
	f := make([][2][3]int64, n)
	for i := range f {
		for j := range f[i] {
			for k := range f[i][j] {
				f[i][j][k] = -1
			}
		}
	}
	var dfs func(int, int, int) int64
	dfs = func(i, j, k int) int64 {
		if i >= n {
			return 0
		}
		if f[i][j][k] != -1 {
			return f[i][j][k]
		}
		x := int64(lane1[i])
		if j == 1 {
			x = int64(lane2[i])
		}
		ans := max(x, dfs(i+1, j, k)+x)
		if k > 0 {
			ans = max(ans, dfs(i+1, j^1, k-1)+x)
			ans = max(ans, dfs(i, j^1, k-1))
		}
		f[i][j][k] = ans
		return ans
	}
	ans := int64(-1e18)
	for i := range lane1 {
		ans = max(ans, dfs(i, 0, 2))
	}
	return ans
}

function maxCoins(lane1: number[], lane2: number[]): number {
    const n = lane1.length;
    const NEG_INF = -1e18;
    const f: number[][][] = Array.from({ length: n }, () =>
        Array.from({ length: 2 }, () => Array(3).fill(NEG_INF)),
    );
    const dfs = (dfs: Function, i: number, j: number, k: number): number => {
        if (i >= n) {
            return 0;
        }
        if (f[i][j][k] !== NEG_INF) {
            return f[i][j][k];
        }
        const x = j === 0 ? lane1[i] : lane2[i];
        let ans = Math.max(x, dfs(dfs, i + 1, j, k) + x);
        if (k > 0) {
            ans = Math.max(ans, dfs(dfs, i + 1, j ^ 1, k - 1) + x);
            ans = Math.max(ans, dfs(dfs, i, j ^ 1, k - 1));
        }
        f[i][j][k] = ans;
        return ans;
    };
    let ans = NEG_INF;
    for (let i = 0; i < n; ++i) {
        ans = Math.max(ans, dfs(dfs, i, 0, 2));
    }
    return ans;
}

3466 - Maximum Coin Collection

3466. Maximum Coin Collection 🔒

Description

Solutions

Solution 1: Memoized Search

All Problems

All Solutions

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3466. Maximum Coin Collection 🔒

Description

Solutions

Solution 1: Memoized Search

All Problems

All Solutions