2948. Make Lexicographically Smallest Array by Swapping Elements LeetCode Solution
In this guide, you will get 2948. Make Lexicographically Smallest Array by Swapping Elements LeetCode Solution with the best time and space complexity. The solution to Make Lexicographically Smallest Array by Swapping Elements problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Make Lexicographically Smallest Array by Swapping Elements
You are given a 0-indexed array of positive integers nums and a positive integer limit.
In one operation, you can choose any two indices i and j and swap nums[i] and nums[j] if |nums[i] – nums[j]| <= limit.
Return the lexicographically smallest array that can be obtained by performing the operation any number of times.
An array a is lexicographically smaller than an array b if in the first position where a and b differ, array a has an element that is less than the corresponding element in b. For example, the array [2,10,3] is lexicographically smaller than the array [10,2,3] because they differ at index 0 and 2 < 10.
Example 1:
Input: nums = [1,5,3,9,8], limit = 2
Output: [1,3,5,8,9]
Explanation: Apply the operation 2 times:
– Swap nums[1] with nums[2]. The array becomes [1,3,5,9,8]
– Swap nums[3] with nums[4]. The array becomes [1,3,5,8,9]
We cannot obtain a lexicographically smaller array by applying any more operations.
Note that it may be possible to get the same result by doing different operations.
Input: nums = [1,7,6,18,2,1], limit = 3
Output: [1,6,7,18,1,2]
Explanation: Apply the operation 3 times:
– Swap nums[1] with nums[2]. The array becomes [1,6,7,18,2,1]
– Swap nums[0] with nums[4]. The array becomes [2,6,7,18,1,1]
– Swap nums[0] with nums[5]. The array becomes [1,6,7,18,1,2]
We cannot obtain a lexicographically smaller array by applying any more operations.
Example 3:
Input: nums = [1,7,28,19,10], limit = 3
Output: [1,7,28,19,10]
Explanation: [1,7,28,19,10] is the lexicographically smallest array we can obtain because we cannot apply the operation on any two indices.
2948. Make Lexicographically Smallest Array by Swapping Elements LeetCode Solution in C++
class Solution {
public:
vector<int> lexicographicallySmallestArray(vector<int>& nums, int limit) {
vector<int> ans(nums.size());
// [[(num, index)]], where the difference between in each pair in each
// `[(num, index)]` group <= `limit`
vector<vector<pair<int, int>>> numAndIndexesGroups;
for (const pair<int, int>& numAndIndex : getNumAndIndexes(nums))
if (numAndIndexesGroups.empty() ||
numAndIndex.first - numAndIndexesGroups.back().back().first > limit) {
// Start a new group.
numAndIndexesGroups.push_back({numAndIndex});
} else {
// Append to the existing group.
numAndIndexesGroups.back().push_back(numAndIndex);
}
for (const vector<pair<int, int>>& numAndIndexesGroup :
numAndIndexesGroups) {
vector<int> sortedNums;
vector<int> sortedIndices;
for (const auto& [num, index] : numAndIndexesGroup) {
sortedNums.push_back(num);
sortedIndices.push_back(index);
}
ranges::sort(sortedIndices);
for (int i = 0; i < sortedNums.size(); ++i)
ans[sortedIndices[i]] = sortedNums[i];
}
return ans;
}
private:
vector<pair<int, int>> getNumAndIndexes(const vector<int>& nums) {
vector<pair<int, int>> numAndIndexes;
for (int i = 0; i < nums.size(); ++i)
numAndIndexes.emplace_back(nums[i], i);
ranges::sort(numAndIndexes);
return numAndIndexes;
}
}; /* code provided by PROGIEZ */
2948. Make Lexicographically Smallest Array by Swapping Elements LeetCode Solution in Java
class Solution {
public int[] lexicographicallySmallestArray(int[] nums, int limit) {
int[] ans = new int[nums.length];
List<List<Pair<Integer, Integer>>> numAndIndexesGroups = new ArrayList<>();
for (Pair<Integer, Integer> numAndIndex : getNumAndIndexes(nums))
if (numAndIndexesGroups.isEmpty() ||
numAndIndex.getKey() -
numAndIndexesGroups.get(numAndIndexesGroups.size() - 1)
.get(numAndIndexesGroups.get(numAndIndexesGroups.size() - 1).size() - 1)
.getKey() >
limit) {
// Start a new group.
numAndIndexesGroups.add(new ArrayList<>(List.of(numAndIndex)));
} else {
// Append to the existing group.
numAndIndexesGroups.get(numAndIndexesGroups.size() - 1).add(numAndIndex);
}
for (List<Pair<Integer, Integer>> numAndIndexesGroup : numAndIndexesGroups) {
List<Integer> sortedNums = new ArrayList<>();
List<Integer> sortedIndices = new ArrayList<>();
for (Pair<Integer, Integer> pair : numAndIndexesGroup) {
sortedNums.add(pair.getKey());
sortedIndices.add(pair.getValue());
}
sortedIndices.sort(null);
for (int i = 0; i < sortedNums.size(); ++i) {
ans[sortedIndices.get(i)] = sortedNums.get(i);
}
}
return ans;
}
private Pair<Integer, Integer>[] getNumAndIndexes(int[] nums) {
Pair<Integer, Integer>[] numAndIndexes = new Pair[nums.length];
for (int i = 0; i < nums.length; ++i)
numAndIndexes[i] = new Pair<>(nums[i], i);
Arrays.sort(numAndIndexes, Comparator.comparing(Pair::getKey));
return numAndIndexes;
}
} // code provided by PROGIEZ
2948. Make Lexicographically Smallest Array by Swapping Elements LeetCode Solution in Python
class Solution:
def lexicographicallySmallestArray(
self,
nums: list[int],
limit: int,
) -> list[int]:
ans = [0] * len(nums)
numAndIndexes = sorted([(num, i) for i, num in enumerate(nums)])
# [[(num, index)]], where the difference between in each pair in each
# `[(num, index)]` group <= `limit`
numAndIndexesGroups: list[list[tuple[int, int]]] = []
for numAndIndex in numAndIndexes:
if (not numAndIndexesGroups or
numAndIndex[0] - numAndIndexesGroups[-1][-1][0] > limit):
# Start a new group.
numAndIndexesGroups.append([numAndIndex])
else:
# Append to the existing group.
numAndIndexesGroups[-1].append(numAndIndex)
for numAndIndexesGroup in numAndIndexesGroups:
sortedNums = [num for num, _ in numAndIndexesGroup]
sortedIndices = sorted([index for _, index in numAndIndexesGroup])
for num, index in zip(sortedNums, sortedIndices):
ans[index] = num
return ans # code by PROGIEZ
