
Problem Statement
Given an array of integers matchsticks
, each element representing the length of a stick, your task is to determine whether you can use all these matchsticks to form a square. Each stick in the array can only be used once and must not be broken. The challenge is to ascertain whether the given lengths can be rearranged to create a square where all sides are of equal length. You are required to return true
if such a square formation is possible with the given matchsticks and false
otherwise.
Examples
Example 1
Input:
matchsticks = [1,1,2,2,2]
Output:
true
Explanation:
You can form a square with length 2, one side of the square came two sticks with length 1.
Example 2
Input:
matchsticks = [3,3,3,3,4]
Output:
false
Explanation:
You cannot find a way to form a square with all the matchsticks.
Constraints
1 <= matchsticks.length <= 15
1 <= matchsticks[i] <= 108
Approach and Intuition
To approach this problem, we need to understand the basic properties of a square and the constraints provided:
- The total length of all matchsticks needs to be divisible by 4 because each side of the square will need to be of equal length. If the total length isn't divisible by 4, it's impossible to form a square.
- Since the forming of a square requires equal division of the matchstick lengths into four equal sums, a possible solution method could involve sorting the matchsticks and trying to distribute them into four groups with equal sums.
- Given the constraint of having at most 15 matchsticks, the brute force approach of trying every possible combination is feasible. However, smarter methods like backtracking can be more efficient:
- Use a depth-first search (DFS) approach to try and build each side of the square by adding matchsticks one by one.
- Use an array
sides
of length 4 to store the current sum of lengths in each side of the square. - Start by trying to place the longest matchstick in any of the four sides and then recursively attempt to place the next ones.
- Backtracking would be required; if adding a stick to a side doesn't eventually lead to a solution, you would remove it and try the next possibility.
- To speed up the process, sorting the matchsticks in descending order might help because placing the longer sticks earlier can help in reaching the sum faster or failing fast if it's not possible.
- Consider the situation where the number of matchsticks is less than four, or the individual lengths are too large compared to others; these are quick checks or early exits in your algorithm.
Through the examples given:
- Example 1: The total length of matchsticks is 8, which is divisible by 4. It is possible to form one side of the square with two sticks of length 1 (totaling 2) and using three sticks of length 2 (one for each of the other sides). Hence, the return is true.
- Example 2: The total length is 16, divisible by 4, but you can't equally distribute the sticks to four sides each of length 4 in any configuration, hence false.
These intuitive steps and checks help guide the development of an algorithm to decide if forming a square is possible with the given matchsticks.
Solutions
- Java
import java.util.HashMap;
class MatchstickSquare {
// Hash map for storing results of recursive checks
public HashMap<Pair<Integer, Integer>, Boolean> dpCache;
// List of all matchsticks
public int[] matchstickLengths;
// The length of one side of the square given total length of sticks
public int targetSideLength;
// Constructor to initialize the memoization
public MatchstickSquare() {
this.dpCache = new HashMap<>();
}
// Method to perform dynamic programming recursion
public boolean dp(Integer currentMask, Integer completedSides) {
int accumulatedLength = 0;
int N = this.matchstickLengths.length;
// Computing hash map key from current state
Pair<Integer, Integer> dpKey = new Pair(currentMask, completedSides);
// Calculate total length of used matchsticks
for(int index = 0; index < N; index++) {
if ((currentMask & (1 << index)) == 0) {
accumulatedLength += this.matchstickLengths[N - 1 - index];
}
}
// Check if a full side of the square has formed
if (accumulatedLength > 0 && accumulatedLength % this.targetSideLength == 0) {
completedSides++;
}
// Successful condition: three complete sides
if (completedSides == 3) {
return true;
}
// Fetch cached results
if (this.dpCache.containsKey(dpKey)) {
return this.dpCache.get(dpKey);
}
boolean result = false;
int currentSideMultiplier = accumulatedLength / this.targetSideLength;
// Remaining length that can be added to current forming side
int remainingSideLength = this.targetSideLength * (currentSideMultiplier + 1) - accumulatedLength;
// Iterate to find next possible matchstick to use
for(int index = 0; index < N; index++) {
if (this.matchstickLengths[N - 1 - index] <= remainingSideLength && (currentMask & (1 << index)) > 0) {
if (dp(currentMask ^ (1 << index), completedSides)) {
result = true;
break;
}
}
}
// Store result in cache
this.dpCache.put(dpKey, result);
return result;
}
// Wrapper function to initiate the matching process
public boolean canFormSquare(int[] matchsticks) {
// Handle the scenario of empty input
if (matchsticks == null || matchsticks.length == 0) {
return false;
}
// Calculate total length and possible side length
int N = matchsticks.length;
int totalLength = 0;
for(int i = 0; i < N; i++) {
totalLength += matchsticks[i];
}
int side = totalLength / 4;
if (side * 4 != totalLength) {
return false;
}
this.matchstickLengths = matchsticks;
this.targetSideLength = side;
return this.dp((1 << N) - 1, 0);
}
}
Analyze and implement a solution to determine if it is possible to arrange a given set of matchsticks to form a square. The code efficiently leverages dynamic programming coupled with bit manipulation to handle this problem.
Instance Variables and Initialization:
- A HashMap (
dpCache
) memorizes intermediate results for subsets of matchsticks. - An array (
matchstickLengths
) stores the lengths of the individual matchsticks. - An integer (
targetSideLength
) represents the length of each side of the target square.
- A HashMap (
Dynamic Programming Approach:
dp(Integer currentMask, Integer completedSides)
: This recursive method checks if forming a square is feasible by assessing each combination of matchsticks (represented bycurrentMask
) and tracking how many sides of the square have been completed (completedSides
).- It employs a bitwise mask (using
currentMask
) to track which matchsticks have been used, with bit positions corresponding to indices in the matchstick array. - As it recurses, it keeps a tally of the total length of the matchsticks used in the current combination, updating
completedSides
when one side of the square completes.
Calculations and Optimizations:
- If the sum of all present matchsticks can't be evenly divided by 4, forming a square is impossible (
canFormSquare
method) - Uses memoization to cache results of subproblems, preventing redundant computations and enhancing efficiency.
- The main function (
canFormSquare
) initializes variables and triggers the recursive depth-first search throughdp
.
- If the sum of all present matchsticks can't be evenly divided by 4, forming a square is impossible (
The recursive method checks feasible combinations, while intermediate checks and the early stopping condition (when three sides are complete) ensure that the algorithm remains efficient. Note that edge cases, such as an empty input array, are handled upfront in the canFormSquare
method. This comprehensive setup effectively addresses the problem, thereby making the solution robust and adaptable to varying lengths of matchsticks.
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