1899. Merge Triplets to Form Target Triplet
Problem Description
In this problem, your goal is to determine if it is possible to achieve a specific target triplet by applying an operation on a given list of triplets. Each triplet consists of three integers. The operation you can perform involves selecting any two triplets and updating one of them by taking the maximum value for each position to form a new triplet.
Given:
- A 2D integer array
triplets
, wheretriplets[i] = [a_i, b_i, c_i]
corresponds to the i-th triplet. - An integer array
target = [x, y, z]
, which is the triplet you are trying to create by using the operation ontriplets
.
The operation you can use is as follows: Choose two different triplets i
and j
(0-indexed) and update triplet j
to be [max(a_i, a_j), max(b_i, b_j), max(c_i, c_j)]
.
The question asks you to return true
if it is possible to have the target triplet [x, y, z] as an element of the triplets
array after applying the operation any number of times (including zero times), or false
otherwise.
Intuition
To solve this problem, the key idea is to realize that each value in the target triplet [x, y, z] must be obtainable by using the max operation within the constraints of the given triplets. To figure this out, we can initialize three variables, d, e, and f, to represent the best candidates for each position of the triplet that we can achieve through the operation.
For each triplet [a, b, c] in triplets
, we check if it is "valid" to use this triplet in the operation towards reaching the target. A valid triplet is one where a is less than or equal to x, b is less than or equal to y, and c is less than or equal to z, meaning that this triplet could potentially be used to reach the target without exceeding it.
If a triplet is valid, we update our best candidates (d, e, f) to be the maximum values seen so far that do not exceed the corresponding target values. At the end of this process, if our best candidate triplet [d, e, f] is equal to the target triplet, then we know it is possible to achieve the target triplet; otherwise, it is not possible.
The key insight is that we don't need to consider the actual sequence of operations. We only need to find the highest values (up to the target values) obtainable for each element of the triplet. If any required value cannot be met or exceeded, the target cannot be reached.
Learn more about Greedy patterns.
Solution Approach
The solution utilizes a simple but effective approach leveraging the idea of maintainability and upgradability in the context of triplets with respect to our target
. We iterate through all the given triplets, updating our candidate triplet [d, e, f]
to the best version that could possibly match our target
.
Here's the step-by-step breakdown of the solution:
-
Initialize three integers
d, e, f
to represent the maximum values we can achieve for each element in the target triplet[x, y, z]
without ever exceeding those values. -
Iterate through each triplet
[a, b, c]
in giventriplets
array. -
For each triplet, check whether it's safe to consider it in the progression towards the target:
a. To be considerate safe,
a
must be less than or equal tox
,b
must be less than or equal toy
, andc
must be less than or equal toz
.b. If the triplet doesn't conform to these conditions, we discard it as it would take us away from our target by exceeding the intended value in at least one position.
-
If the triplet
[a, b, c]
is valid, we take the maximum values between our current candidates[d, e, f]
and[a, b, c]
. This basically simulates the allowed operation but only in the direction of increasing our chances of reaching the target without going past it:a. Update
d
withmax(d, a)
b. Update
e
withmax(e, b)
c. Update
f
withmax(f, c)
-
After iterating through all triplets, check if the candidate triplet
[d, e, f]
matches thetarget
:a. If they match, it implies that we can indeed form the target by potentially merging the triplets without ever needing to exceed the target values.
b. If they don't match, we conclude that it is not possible to reach the target given the operations and constraints defined.
In this solution, we essentially simulate the process of using the maximum operation to gradually upgrade a fictional triplet starting from [0, 0, 0]
to the maximum values that it can reach without exceeding the target values [x, y, z]
. If at the end of this process, we have successfully reached [x, y, z]
, it means that the operations can indeed reconstruct the target triplet from the given array of triplets.
No additional data structures or patterns are required for this approach, as we simply use loop iteration and conditionals to handle the core logic, and rely on basic variable assignments for state tracking.
Ready to land your dream job?
Unlock your dream job with a 2-minute evaluator for a personalized learning plan!
Start EvaluatorExample Walkthrough
Consider an example where the triplets
array is [[3,4,5], [1,2,3], [2,3,4]]
and the target
triplet is [2,3,5]
. Let's use the solution approach to check if we can achieve the target
.
-
We initialize
d, e, f
to0
since we start out with a fictional triplet[0, 0, 0]
. -
Start iterating through the
triplets
array:- First triplet:
[3,4,5]
- Since
3
is greater than2
(our x value intarget
), this triplet cannot contribute to forming the first element of thetarget
. We do not update ourd, e, f
values.
- Since
- Second triplet:
[1,2,3]
- All elements are less than or equal to the corresponding
target
elements. We can consider this entire triplet. d = max(d, 1)
→d = 1
e = max(e, 2)
→e = 2
f = max(f, 3)
→f = 3
- All elements are less than or equal to the corresponding
- Third triplet:
[2,3,4]
- All elements are less than or equal to the corresponding
target
elements. We can consider this entire triplet as well. d = max(d, 2)
→d = 2
e = max(e, 3)
→e = 3
f = max(f, 4)
→f = 4
(Note thatf
is not equal to5
, which is the target's third value)
- All elements are less than or equal to the corresponding
- First triplet:
-
After iterating through all triplets, our constructed candidate triplet
[d, e, f]
is[2,3,4]
. -
Finally, we compare our candidate
[2,3,4]
with thetarget
[2,3,5]
. Since the third element does not match (4
instead of5
), we determine that it is not possible to achieve thetarget
triplet[2,3,5]
.
In this example walk through, we clearly see how the solution methodically processes each triplet in relation to the target
and upgrades the candidate triplet [d, e, f]
only in ways that are safe and in accordance with our goal of reaching the target
. Despite the operations we carried out, the target's third value was not met, leading us to the conclusion that under the given constraints, forming the target is not possible.
Solution Implementation
1from typing import List
2
3class Solution:
4 def mergeTriplets(self, triplets: List[List[int]], target: List[int]) -> bool:
5 # Initialize variables to track the maximum values for each position
6 max_x, max_y, max_z = 0, 0, 0
7
8 # Unpack target values into separate variables for clarity
9 target_x, target_y, target_z = target
10
11 # Iterate through each triplet in the list of triplets
12 for triplet in triplets:
13 # Unpack the values in the current triplet
14 a, b, c = triplet
15
16 # Check if the current triplet is valid by comparing it to the target
17 if a <= target_x and b <= target_y and c <= target_z:
18 # Update the maximum values seen so far for each position, if applicable
19 max_x = max(max_x, a)
20 max_y = max(max_y, b)
21 max_z = max(max_z, c)
22
23 # Return True if the maximum values match the target values, otherwise return False
24 return [max_x, max_y, max_z] == target
25
1class Solution {
2 public boolean mergeTriplets(int[][] triplets, int[] target) {
3 // Extract the target values for easy reference
4 int targetX = target[0];
5 int targetY = target[1];
6 int targetZ = target[2];
7
8 // Initialize the max values found in the triplets
9 int maxX = 0;
10 int maxY = 0;
11 int maxZ = 0;
12
13 // Iterate over each triplet in the given array of triplets
14 for (int[] triplet : triplets) {
15 // Extract the values of the current triplet
16 int currentX = triplet[0];
17 int currentY = triplet[1];
18 int currentZ = triplet[2];
19
20 // Check if the current triplet can potentially contribute to
21 // forming the target triplet without exceeding any target value
22 if (currentX <= targetX && currentY <= targetY && currentZ <= targetZ) {
23 // Update the maximum values found that do not exceed the target values
24 maxX = Math.max(maxX, currentX);
25 maxY = Math.max(maxY, currentY);
26 maxZ = Math.max(maxZ, currentZ);
27 }
28 }
29
30 // Return true if the maximum values found match the target values
31 return maxX == targetX && maxY == targetY && maxZ == targetZ;
32 }
33}
34
1class Solution {
2public:
3 bool mergeTriplets(vector<vector<int>>& triplets, vector<int>& target) {
4 // Extract the target values for comparison.
5 int targetX = target[0], targetY = target[1], targetZ = target[2];
6
7 // Initialize variables to keep track of the maximum values of the triplets.
8 int maxX = 0, maxY = 0, maxZ = 0;
9
10 // Iterate over each triplet in the input list.
11 for (auto& triplet : triplets) {
12 // Extract the triplet values for comparison.
13 int currentX = triplet[0], currentY = triplet[1], currentZ = triplet[2];
14
15 // Check if the current triplet's values are less than or equal
16 // to the corresponding target values.
17 if (currentX <= targetX && currentY <= targetY && currentZ <= targetZ) {
18 // Update the running maxima if the current values are greater.
19 maxX = max(maxX, currentX);
20 maxY = max(maxY, currentY);
21 maxZ = max(maxZ, currentZ);
22 }
23 }
24
25 // Determine if the largest triplets found match the target triplet.
26 return maxX == targetX && maxY == targetY && maxZ == targetZ;
27 }
28};
29
1/**
2 * Checks whether it is possible to form a target triplet from a set of given triplets.
3 *
4 * @param {number[][]} triplets - An array containing triplets (arrays of three numbers).
5 * @param {number[]} target - An array of three integers representing the target triplet.
6 * @returns {boolean} - True if the target triplet can be formed, otherwise false.
7 */
8function mergeTriplets(triplets: number[][], target: number[]): boolean {
9 // Destructure the target triplet into the required values.
10 const [targetX, targetY, targetZ] = target;
11
12 // Initialize max values for x, y, z to 0.
13 let [maxX, maxY, maxZ] = [0, 0, 0];
14
15 // Iterate over each triplet in the given array.
16 for (const [tripletX, tripletY, tripletZ] of triplets) {
17 // Check if the values of the current triplet are less than or equal to the target values.
18 if (tripletX <= targetX && tripletY <= targetY && tripletZ <= targetZ) {
19 // Update max values, if the current value is higher than what was seen before.
20 maxX = Math.max(maxX, tripletX);
21 maxY = Math.max(maxY, tripletY);
22 maxZ = Math.max(maxZ, tripletZ);
23 }
24 }
25
26 // Return true if the max values for x, y, z match the target triplet exactly.
27 return maxX === targetX && maxY === targetY && maxZ === targetZ;
28}
29
Time and Space Complexity
The time complexity of the code is O(n)
, where n
is the number of triplets. This is because there is a single loop that iterates over each element in the triplets list once, and within this loop, the operations performed are all constant time operations (comparisons, assignments, and max()
function on integers).
The space complexity of the code is O(1)
, independent of the number of triplets. The only extra space used is for the variables d
, e
, and f
, which hold the maximum values of the first, second, and third elements of the triplets that can be part of the merge to reach the target. The variables x
, y
, and z
are just references to the elements of the target
list and do not count towards additional space.
Learn more about how to find time and space complexity quickly using problem constraints.
How does quick sort divide the problem into subproblems?
Recommended Readings
Greedy Introduction div class responsive iframe iframe src https www youtube com embed WTslqPbj7I title YouTube video player frameborder 0 allow accelerometer autoplay clipboard write encrypted media gyroscope picture in picture web share allowfullscreen iframe div When do we use greedy Greedy algorithms tend to solve optimization problems Typically they will ask you to calculate the max min of some value Commonly you may see this phrased in the problem as max min longest shortest largest smallest etc These keywords can be identified by just scanning
LeetCode Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way we
Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https algomonster s3 us east 2 amazonaws com recursion jpg You first