Minimum Score Triangulation of Polygon

🎯

Minimum Score Triangulation of Polygon

mediumDPGoogle

Imagine you want to cut a convex polygon into triangles such that the sum of the triangle scores (product of vertices) is minimized. This problem models optimizing costs in polygon triangulation, useful in graphics and computational geometry.

💡 This problem is a classic example of interval dynamic programming where the solution depends on dividing the problem into smaller subproblems defined by intervals. Beginners often struggle because the problem requires understanding how to break down the polygon into smaller parts and combine results optimally.

📋

Problem Statement

Given an array values[] of integers representing the vertices of a convex polygon in order, find the minimum possible sum of the scores of triangulations of the polygon. The score of a triangle formed by vertices i, j, k is values[i] * values[j] * values[k]. Return the minimum score possible by triangulating the polygon.

3 ≤ n = values.length ≤ 501 ≤ values[i] ≤ 100

💡

Example

Input"values = [1, 2, 3]"

Output6

Only one triangle possible with vertices 1,2,3. Score = 1*2*3 = 6.

Input"values = [3, 7, 4, 5]"

Output144

Optimal triangulation splits polygon into triangles with minimal sum of products.

• Minimum polygon size (3 vertices) → output is product of those three vertices
• All vertices have the same value → output is predictable product times number of triangles
• Polygon with increasing vertex values → tests if algorithm finds minimal triangulation
• Polygon with maximum allowed vertex values → tests efficiency and correctness under constraints

🔁

Why DP?

Why greedy fails:

A greedy approach that always picks the triangle with the smallest immediate product fails because it ignores future costs. For example, picking a small triangle early may force large costly triangles later, increasing total score.

DP state:

dp[i][j] represents the minimum triangulation score for the polygon formed by vertices from index i to j inclusive.

Recurrence:dp[i][j] = min over k in (i+1 to j-1) of (dp[i][k] + dp[k][j] + values[i]*values[k]*values[j])

For each interval [i,j], try all possible k to split the polygon into two smaller polygons and add the cost of the triangle formed by i,k,j. Choose the minimal sum.

⚠️

Common Mistakes

Forgetting base case when interval length < 3

Recursion or DP accesses invalid indices or returns wrong results

✅ Add base case returning 0 when j <= i + 1

Not initializing dp[i][j] to infinity before minimization

dp[i][j] may hold incorrect default value, leading to wrong minimal cost

✅ Set dp[i][j] = float('inf') or Integer.MAX_VALUE before inner loop

Mixing up indices i, j, k in loops

Incorrect triangle cost calculation or out-of-bound errors

✅ Carefully use i < k < j and ensure loops respect these bounds

Using greedy approach to pick smallest triangle first

Leads to suboptimal total triangulation cost

✅ Use DP to consider all splits and pick global minimum

Not caching results in recursion leading to TLE

Exponential time complexity and timeout in large inputs

✅ Implement memoization or bottom-up DP

🧠

Brute Force (Pure Recursion)

💡 Starting with brute force helps understand the problem's recursive structure and why naive recursion is inefficient due to repeated computations.

Intuition

Try every possible way to split the polygon into two smaller polygons by choosing a vertex k between i and j, recursively compute minimal triangulation for each sub-polygon, and add the cost of the triangle formed by i, k, j.

Algorithm

1. Define a recursive function minScore(i, j) that returns minimal triangulation score for polygon vertices i to j.
2. If the polygon has less than 3 vertices (j <= i+1), return 0 since no triangle can be formed.
3. For each vertex k between i and j, recursively compute minScore(i, k) and minScore(k, j), then add the cost of triangle i,k,j.
4. Return the minimum sum over all k.

💡 The recursion explores all ways to split the polygon, which is conceptually simple but leads to exponential calls.

Recurrence:dp[i][j] = min_{i<k<j} (dp[i][k] + dp[k][j] + values[i]*values[k]*values[j])

</>

Code

from sys import setrecursionlimit
setrecursionlimit(10**7)

def minScoreTriangulation(values):
    n = len(values)
    def dfs(i, j):
        if j <= i + 1:
            return 0
        res = float('inf')
        for k in range(i + 1, j):
            cost = dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j]
            if cost < res:
                res = cost
        return res
    return dfs(0, n - 1)

# Driver code
if __name__ == '__main__':
    values = [1, 3, 1, 4, 1, 5]
    print(minScoreTriangulation(values))  # Expected output: 13

Line Notes

def dfs(i, j):Defines recursive function to compute minimal triangulation score for interval [i,j]

if j <= i + 1:Base case: no triangle possible if interval length less than 3

for k in range(i + 1, j):Try all possible vertices k to split polygon into two parts

cost = dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j]Calculate cost of triangulation including triangle i,k,j

return dfs(0, n - 1)Start recursion for entire polygon

public class Solution {
    public int minScoreTriangulation(int[] values) {
        int n = values.length;
        return dfs(values, 0, n - 1);
    }
    private int dfs(int[] values, int i, int j) {
        if (j <= i + 1) return 0;
        int res = Integer.MAX_VALUE;
        for (int k = i + 1; k < j; k++) {
            int cost = dfs(values, i, k) + dfs(values, k, j) + values[i] * values[k] * values[j];
            if (cost < res) res = cost;
        }
        return res;
    }
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] values = {1, 3, 1, 4, 1, 5};
        System.out.println(sol.minScoreTriangulation(values)); // Expected: 13
    }
}

Line Notes

public int minScoreTriangulation(int[] values)Entry point to start recursion for entire polygon

if (j <= i + 1) return 0;Base case: no triangle if less than 3 vertices

for (int k = i + 1; k < j; k++)Try all possible splits between i and j

int cost = dfs(values, i, k) + dfs(values, k, j) + values[i] * values[k] * values[j];Calculate cost including triangle i,k,j

return res;Return minimal triangulation cost for interval

#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int dfs(const vector<int>& values, int i, int j) {
    if (j <= i + 1) return 0;
    int res = INT_MAX;
    for (int k = i + 1; k < j; ++k) {
        int cost = dfs(values, i, k) + dfs(values, k, j) + values[i] * values[k] * values[j];
        if (cost < res) res = cost;
    }
    return res;
}

int minScoreTriangulation(vector<int>& values) {
    return dfs(values, 0, (int)values.size() - 1);
}

int main() {
    vector<int> values = {1, 3, 1, 4, 1, 5};
    cout << minScoreTriangulation(values) << endl; // Expected: 13
    return 0;
}

Line Notes

int dfs(const vector<int>& values, int i, int j)Recursive function for minimal triangulation score in interval [i,j]

if (j <= i + 1) return 0;Base case: no triangle if interval too small

for (int k = i + 1; k < j; ++k)Try all possible vertices k to split polygon

int cost = dfs(values, i, k) + dfs(values, k, j) + values[i] * values[k] * values[j];Calculate cost including triangle i,k,j

return dfs(values, 0, (int)values.size() - 1);Start recursion for entire polygon

function minScoreTriangulation(values) {
    const n = values.length;
    function dfs(i, j) {
        if (j <= i + 1) return 0;
        let res = Infinity;
        for (let k = i + 1; k < j; k++) {
            const cost = dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j];
            if (cost < res) res = cost;
        }
        return res;
    }
    return dfs(0, n - 1);
}

// Test
console.log(minScoreTriangulation([1, 3, 1, 4, 1, 5])); // Expected: 13

Line Notes

function dfs(i, j)Recursive helper to compute minimal triangulation for interval [i,j]

if (j <= i + 1) return 0;Base case: no triangle if interval too small

for (let k = i + 1; k < j; k++)Try all possible splits between i and j

const cost = dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j];Calculate cost including triangle i,k,j

return dfs(0, n - 1);Start recursion for entire polygon

⏱

Complexity

TimeO(n^3)

SpaceO(n) due to recursion stack

For each interval [i,j], we try all k between i and j, leading to O(n^3) calls in total.

💡 For n=20, this means about 8000 operations, which is borderline but still slow without memoization.

Interview Verdict: TLE for large inputs

This approach is too slow for large inputs but is essential to understand the problem's recursive structure.

🧠

Top-Down DP with Memoization

💡 Memoization avoids recomputing overlapping subproblems by caching results, drastically improving efficiency while keeping the recursive structure.

Intuition

Use the same recursive approach but store results of dp[i][j] after computing them once, so repeated calls return cached results instantly.

Algorithm

1. Initialize a 2D dp array with -1 to indicate uncomputed states.
2. Define recursive function dfs(i, j) that returns dp[i][j] if computed.
3. If not computed, compute minimal triangulation by trying all k between i and j.
4. Store and return the computed dp[i][j].

💡 Memoization adds caching to recursion, which is conceptually simple but requires careful indexing.

Recurrence:dp[i][j] = min_{i<k<j} (dp[i][k] + dp[k][j] + values[i]*values[k]*values[j])

</>

Code

def minScoreTriangulation(values):
    n = len(values)
    dp = [[-1] * n for _ in range(n)]
    def dfs(i, j):
        if j <= i + 1:
            return 0
        if dp[i][j] != -1:
            return dp[i][j]
        res = float('inf')
        for k in range(i + 1, j):
            cost = dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j]
            if cost < res:
                res = cost
        dp[i][j] = res
        return res
    return dfs(0, n - 1)

# Driver code
if __name__ == '__main__':
    values = [1, 3, 1, 4, 1, 5]
    print(minScoreTriangulation(values))  # Expected output: 13

Line Notes

dp = [[-1] * n for _ in range(n)]Initialize dp table with -1 to mark uncomputed states

if dp[i][j] != -1:Return cached result if already computed

dp[i][j] = resStore computed minimal triangulation cost for interval [i,j]

return dfs(0, n - 1)Start recursion with memoization for entire polygon

public class Solution {
    private int[][] dp;
    public int minScoreTriangulation(int[] values) {
        int n = values.length;
        dp = new int[n][n];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                dp[i][j] = -1;
        return dfs(values, 0, n - 1);
    }
    private int dfs(int[] values, int i, int j) {
        if (j <= i + 1) return 0;
        if (dp[i][j] != -1) return dp[i][j];
        int res = Integer.MAX_VALUE;
        for (int k = i + 1; k < j; k++) {
            int cost = dfs(values, i, k) + dfs(values, k, j) + values[i] * values[k] * values[j];
            if (cost < res) res = cost;
        }
        dp[i][j] = res;
        return res;
    }
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] values = {1, 3, 1, 4, 1, 5};
        System.out.println(sol.minScoreTriangulation(values)); // Expected: 13
    }
}

Line Notes

dp = new int[n][n];Create dp table to cache results

dp[i][j] = -1;Mark states as uncomputed

if (dp[i][j] != -1) return dp[i][j];Return cached result if available

dp[i][j] = res;Store computed minimal triangulation cost

#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int dfs(const vector<int>& values, int i, int j, vector<vector<int>>& dp) {
    if (j <= i + 1) return 0;
    if (dp[i][j] != -1) return dp[i][j];
    int res = INT_MAX;
    for (int k = i + 1; k < j; ++k) {
        int cost = dfs(values, i, k, dp) + dfs(values, k, j, dp) + values[i] * values[k] * values[j];
        if (cost < res) res = cost;
    }
    dp[i][j] = res;
    return res;
}

int minScoreTriangulation(vector<int>& values) {
    int n = (int)values.size();
    vector<vector<int>> dp(n, vector<int>(n, -1));
    return dfs(values, 0, n - 1, dp);
}

int main() {
    vector<int> values = {1, 3, 1, 4, 1, 5};
    cout << minScoreTriangulation(values) << endl; // Expected: 13
    return 0;
}

Line Notes

vector<vector<int>> dp(n, vector<int>(n, -1));Initialize dp table with -1 for memoization

if (dp[i][j] != -1) return dp[i][j];Return cached result if computed

dp[i][j] = res;Cache computed minimal triangulation cost

return dfs(values, 0, n - 1, dp);Start memoized recursion

function minScoreTriangulation(values) {
    const n = values.length;
    const dp = Array.from({ length: n }, () => Array(n).fill(-1));
    function dfs(i, j) {
        if (j <= i + 1) return 0;
        if (dp[i][j] !== -1) return dp[i][j];
        let res = Infinity;
        for (let k = i + 1; k < j; k++) {
            const cost = dfs(i, k) + dfs(k, j) + values[i] * values[k] * values[j];
            if (cost < res) res = cost;
        }
        dp[i][j] = res;
        return res;
    }
    return dfs(0, n - 1);
}

// Test
console.log(minScoreTriangulation([1, 3, 1, 4, 1, 5])); // Expected: 13

Line Notes

const dp = Array.from({ length: n }, () => Array(n).fill(-1));Create 2D dp array initialized to -1 for memoization

if (dp[i][j] !== -1) return dp[i][j];Return cached result if available

dp[i][j] = res;Store computed minimal triangulation cost

return dfs(0, n - 1);Start memoized recursion for entire polygon

⏱

Complexity

TimeO(n^3)

SpaceO(n^2) for dp table + O(n) recursion stack

Memoization ensures each dp[i][j] is computed once, and each computation tries O(n) splits, totaling O(n^3).

💡 For n=50, this is about 125,000 operations, which is efficient enough for interviews.

Interview Verdict: Accepted

Memoization makes the solution efficient and practical for interview constraints.

🧠

Bottom-Up DP (Tabulation)

💡 Tabulation builds the solution iteratively from smaller intervals to larger ones, avoiding recursion and making the process explicit and often easier to debug.

Intuition

Fill a 2D dp table where dp[i][j] stores minimal triangulation score for polygon vertices i to j. Start with small intervals and increase interval length, using previously computed smaller intervals.

Algorithm

1. Initialize dp table with zeros for intervals of length less than 3.
2. Iterate over interval lengths from 3 to n.
3. For each interval [i, j] of current length, compute dp[i][j] by trying all k between i and j.
4. Update dp[i][j] with minimal cost of splitting polygon and adding triangle cost.

💡 This approach explicitly fills the dp table in a safe order ensuring dependencies are computed first.

Recurrence:dp[i][j] = min_{i<k<j} (dp[i][k] + dp[k][j] + values[i]*values[k]*values[j])

</>

Code

def minScoreTriangulation(values):
    n = len(values)
    dp = [[0] * n for _ in range(n)]
    for length in range(3, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            dp[i][j] = float('inf')
            for k in range(i + 1, j):
                cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j]
                if cost < dp[i][j]:
                    dp[i][j] = cost
    return dp[0][n - 1]

# Driver code
if __name__ == '__main__':
    values = [1, 3, 1, 4, 1, 5]
    print(minScoreTriangulation(values))  # Expected output: 13

Line Notes

dp = [[0] * n for _ in range(n)]Initialize dp table with zeros for base cases

for length in range(3, n + 1):Iterate over all possible polygon interval lengths

for i in range(n - length + 1):Start index of interval

dp[i][j] = float('inf')Initialize dp[i][j] to infinity before finding minimum

cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j]Calculate cost of triangulation splitting at k

public class Solution {
    public int minScoreTriangulation(int[] values) {
        int n = values.length;
        int[][] dp = new int[n][n];
        for (int length = 3; length <= n; length++) {
            for (int i = 0; i <= n - length; i++) {
                int j = i + length - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i + 1; k < j; k++) {
                    int cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j];
                    if (cost < dp[i][j]) dp[i][j] = cost;
                }
            }
        }
        return dp[0][n - 1];
    }
    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] values = {1, 3, 1, 4, 1, 5};
        System.out.println(sol.minScoreTriangulation(values)); // Expected: 13
    }
}

Line Notes

int[][] dp = new int[n][n];Create dp table for storing minimal triangulation costs

for (int length = 3; length <= n; length++)Iterate over polygon interval lengths

dp[i][j] = Integer.MAX_VALUE;Initialize dp[i][j] before minimizing

int cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j];Calculate cost for splitting polygon at k

return dp[0][n - 1];Return minimal triangulation cost for entire polygon

#include <iostream>
#include <vector>
#include <climits>
using namespace std;

int minScoreTriangulation(vector<int>& values) {
    int n = (int)values.size();
    vector<vector<int>> dp(n, vector<int>(n, 0));
    for (int length = 3; length <= n; ++length) {
        for (int i = 0; i <= n - length; ++i) {
            int j = i + length - 1;
            dp[i][j] = INT_MAX;
            for (int k = i + 1; k < j; ++k) {
                int cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j];
                if (cost < dp[i][j]) dp[i][j] = cost;
            }
        }
    }
    return dp[0][n - 1];
}

int main() {
    vector<int> values = {1, 3, 1, 4, 1, 5};
    cout << minScoreTriangulation(values) << endl; // Expected: 13
    return 0;
}

Line Notes

vector<vector<int>> dp(n, vector<int>(n, 0));Initialize dp table with zeros for base cases

for (int length = 3; length <= n; ++length)Iterate over polygon interval lengths

dp[i][j] = INT_MAX;Initialize dp[i][j] before minimizing

int cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j];Calculate cost for splitting polygon at k

return dp[0][n - 1];Return minimal triangulation cost for entire polygon

function minScoreTriangulation(values) {
    const n = values.length;
    const dp = Array.from({ length: n }, () => Array(n).fill(0));
    for (let length = 3; length <= n; length++) {
        for (let i = 0; i <= n - length; i++) {
            let j = i + length - 1;
            dp[i][j] = Infinity;
            for (let k = i + 1; k < j; k++) {
                const cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j];
                if (cost < dp[i][j]) dp[i][j] = cost;
            }
        }
    }
    return dp[0][n - 1];
}

// Test
console.log(minScoreTriangulation([1, 3, 1, 4, 1, 5])); // Expected: 13

Line Notes

const dp = Array.from({ length: n }, () => Array(n).fill(0));Initialize dp table with zeros for base cases

for (let length = 3; length <= n; length++)Iterate over polygon interval lengths

dp[i][j] = Infinity;Initialize dp[i][j] before minimizing

const cost = dp[i][k] + dp[k][j] + values[i] * values[k] * values[j];Calculate cost for splitting polygon at k

return dp[0][n - 1];Return minimal triangulation cost for entire polygon

⏱

Complexity

TimeO(n^3)

SpaceO(n^2)

Triple nested loops: interval length, start index, and split point k, each up to n.

💡 For n=50, this is about 125,000 operations, efficient for interview constraints.

Interview Verdict: Accepted

Bottom-up DP is preferred in interviews for clarity and avoiding recursion overhead.

📊

All Approaches - One-Glance Tradeoffs

💡 Memoization or bottom-up DP are the best choices for interviews due to efficiency and clarity.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n^3)	O(n) recursion stack	Yes (deep recursion)	No	Mention only - never code
2. Top-Down DP with Memoization	O(n^3)	O(n^2) + O(n) recursion stack	Possible but less likely	Yes	Good for quick implementation and explanation
3. Bottom-Up DP (Tabulation)	O(n^3)	O(n^2)	No	Yes	Preferred for clarity and iterative coding

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice coding all approaches, and prepare to explain tradeoffs clearly in interviews.

How to Present

Clarify problem constraints and inputs.Explain brute force recursion to show understanding of problem structure.Introduce memoization to optimize recursion.Present bottom-up DP as the optimal iterative solution.Discuss time and space complexities and possible follow-ups.

Time Allocation

Clarify: 3min → Brute Force: 5min → Memoization: 7min → Bottom-Up DP: 7min → Testing & Discussion: 3min. Total ~25min

What the Interviewer Tests

Understanding of interval DP, ability to optimize recursion with memoization, and clarity in explaining DP states and transitions.

Common Follow-ups

• Can you reduce space complexity? → Not easily for this problem due to 2D dependencies.
• What if polygon vertices are not convex? → Problem constraints require convex polygon; otherwise, triangulation cost is undefined.
• Can you reconstruct the triangulation? → Yes, by storing choices of k during DP.
• What if we want maximum score instead? → Change min to max in recurrence.

💡 These follow-ups test deeper understanding of DP, problem constraints, and ability to adapt solutions.

🔍

Pattern Recognition

When to Use

1) Problem involves partitioning intervals or sequences; 2) Optimal substructure with overlapping subproblems; 3) Cost depends on triplets or groups inside intervals; 4) Requires minimizing or maximizing sum over partitions.

Signature Phrases

'minimum score triangulation''convex polygon vertices''product of vertices''partition polygon into triangles'

NOT This Pattern When

Problems involving simple greedy or linear DP without interval partitioning.

Similar Problems

Burst Balloons - similar interval DP with product costsPalindrome Partitioning - interval DP with partitionsMatrix Chain Multiplication - classic interval DP minimizing multiplication cost

Practice

1. You are given an n x n integer matrix. You want to find a path from the top row to the bottom row such that you move one step down each time, and at each step you can move to the same column, the column to the left, or the column to the right. The goal is to minimize the sum of the values along this path. Which algorithmic approach guarantees finding the minimum sum efficiently?

easy

A. Greedy algorithm that picks the minimum adjacent value at each step

B. Sorting each row and picking the smallest values independently

C. Depth-first search exploring all paths without memoization

D. Dynamic programming that builds solutions row by row using previous row results

Solution

1. Step 1: Understand the problem constraints

   The problem requires considering all possible paths moving down and diagonally, which suggests overlapping subproblems and optimal substructure.

2. Step 2: Identify the suitable algorithm

   Dynamic programming fits because it efficiently computes minimum sums for each cell based on the previous row's results, avoiding redundant calculations.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Greedy and sorting fail to consider future steps; DFS without memoization is exponential [OK]

Hint: DP uses previous row results to build solutions [OK]

Common Mistakes:

• Thinking greedy or sorting per row suffices
• Ignoring overlapping subproblems

2. What is the time complexity of the bottom-up dynamic programming solution for the Stone Game problem with n piles, and why?

medium

A. O(n^3) because of three nested loops over the piles

B. O(n^2) because the DP table of size nxn is filled once with constant work per cell

C. O(2^n) because all subsets of piles are considered

D. O(n) because only linear passes over the piles are needed

Solution

1. Step 1: Identify loops in bottom-up DP

   There are two nested loops: one for length from 2 to n, and one for start index i, total O(n^2) iterations.

2. Step 2: Constant work per dp[i][j]

   Each dp[i][j] is computed with a constant number of operations (max of two values), so total time is O(n^2).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   DP table size nxn filled once with O(1) work per cell [OK]

Hint: Two nested loops over intervals -> O(n^2) [OK]

Common Mistakes:

• Confusing recursion stack space with time
• Assuming triple nested loops cause O(n^3)

3. What is the time and space complexity of the optimal space-optimized bottom-up DP solution for the Unique Paths II problem on an m x n grid with obstacles?

medium

A. Time: O(m*n), Space: O(m*n)

B. Time: O(m+n), Space: O(m+n)

C. Time: O(m*n), Space: O(n)

D. Time: O(2^(m+n)), Space: O(m*n)

Solution

1. Step 1: Analyze time complexity

   The algorithm iterates over each cell once in nested loops: O(m*n).

2. Step 2: Analyze space complexity

   Uses a single 1D dp array of length n, so space is O(n).

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Nested loops over m and n, dp array size n [OK]

Hint: Nested loops over m and n, dp array size n [OK]

Common Mistakes:

• Confusing recursion stack space with dp array
• Assuming full 2D dp array used
• Mistaking linear time for exponential

4. The following code attempts to compute the number of unique paths in an m x n grid using a space-optimized DP approach:

def uniquePaths(m, n):
    dp = [1] * n
    for i in range(1, m):
        for j in range(1, n):
            dp[j] = dp[j] + dp[j]
    return dp[-1]

What is the bug in this code?

medium

A. Line 3 should start loop from 0 instead of 1

B. Line 4 incorrectly doubles dp[j] instead of adding dp[j-1]

C. Line 5 should return dp[0] instead of dp[-1]

D. Line 2 initializes dp with wrong size

Solution

1. Step 1: Analyze inner loop update

   Line 4 updates dp[j] by adding dp[j] to itself, doubling the value instead of adding dp[j-1].

2. Step 2: Identify correct update

   The correct update is dp[j] += dp[j - 1] to accumulate paths from left and top cells.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Doubling dp[j] breaks path count logic -> bug at line 4 [OK]

Hint: Check dp update uses dp[j-1], not dp[j] twice [OK]

Common Mistakes:

• Using dp[j] + dp[j] instead of dp[j] + dp[j-1]
• Off-by-one errors in loops

5. Suppose the problem is modified so that the matrix can contain '1's and '0's, but you want to find the largest square containing only '1's where you are allowed to flip at most one '0' inside the square to '1'. Which of the following approaches correctly adapts the DP solution to handle this variant efficiently?

hard

A. Maintain two DP arrays: one for squares with zero flips and one for squares with one flip allowed, updating both simultaneously

B. Use the original DP but treat all '0's as '1's to allow flipping implicitly

C. Run the original DP multiple times, each time flipping a different '0' to '1' and taking the max result

D. Use a greedy approach to find the largest square ignoring zeros, then check if flipping one zero inside is possible

Solution

1. Step 1: Understand the variant

   Allowing one flip means tracking squares formed with zero or one zero flipped inside.

2. Step 2: Adapt DP states

   Maintain two DP states per cell: max square size without flips and with one flip used, updating both based on neighbors.

3. Step 3: Reject naive approaches

   Original DP ignores flips; multiple runs are inefficient; greedy ignores DP dependencies.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two DP arrays track flip states efficiently [OK]

Hint: Track DP states with and without flips [OK]

Common Mistakes:

• Ignoring flip state in DP
• Trying multiple brute force runs
• Using greedy without DP