Approach Summary
Optimal solution for Minimum Number of Moves to Make Palindrome with O(n^2) time complexity and O(1) space complexity.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n^2) | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public void swap(char arr[], int i, int j){
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public int minMovesToMakePalindrome(String s) {
int minSwaps = 0;
int left = 0;
int right = s.length()-1;
char arr[] = s.toCharArray();
while(left < right){
int l = left;
int r = right;
while(arr[l] != arr[r]){
r--;
}
if(l == r){
// No matching character found, swap with the adjacent right character
swap(arr, r, r+1);
minSwaps++;
continue;
}
else{
while(r < right){
// Swap characters to make them same
swap(arr, r, r+1);
minSwaps++;
r++;
}
}
left++;
right--;
}
return minSwaps;
}
}class Solution {
public:
int minMovesToMakePalindrome(string s) {
int minSwaps = 0;
int left = 0;
int right = s.size()-1;
while(left < right){
int l = left;
int r = right;
while(s[l] != s[r]){
r--;
}
if(l == r){
// No matching character found, swap with the adjacent right character
swap(s[r], s[r+1]);
minSwaps++;
continue;
}
else{
// Swap characters to make them same
while(r < right){
swap(s[r], s[r+1]);
minSwaps++;
r++;
}
}
left++;
right--;
}
return minSwaps;
}
};class Solution:
def minimum_number_of_moves_to_make_palindrome(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Minimum Number of Moves to Make Palindrome
*/
function minimumNumberOfMovesToMakePalindrome(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n^2)
Space: O(1)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public void swap(char arr[], int i, int j){
char temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
public int minMovesToMakePalindrome(String s) {
int minSwaps = 0;
int left = 0;
int right = s.length()-1;
char arr[] = s.toCharArray();
while(left < right){
int l = left;
int r = right;
while(arr[l] != arr[r]){
r--;
}
if(l == r){
// No matching character found, swap with the adjacent right character
swap(arr, r, r+1);
minSwaps++;
continue;
}
else{
while(r < right){
// Swap characters to make them same
swap(arr, r, r+1);
minSwaps++;
r++;
}
}
left++;
right--;
}
return minSwaps;
}
}class Solution {
public:
int minMovesToMakePalindrome(string s) {
int minSwaps = 0;
int left = 0;
int right = s.size()-1;
while(left < right){
int l = left;
int r = right;
while(s[l] != s[r]){
r--;
}
if(l == r){
// No matching character found, swap with the adjacent right character
swap(s[r], s[r+1]);
minSwaps++;
continue;
}
else{
// Swap characters to make them same
while(r < right){
swap(s[r], s[r+1]);
minSwaps++;
r++;
}
}
left++;
right--;
}
return minSwaps;
}
};class Solution:
def minimum_number_of_moves_to_make_palindrome(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Minimum Number of Moves to Make Palindrome
*/
function minimumNumberOfMovesToMakePalindrome(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Two Pointers Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Two Pointers pattern.
Framework: Two Pointers • Interview Prep: Tested at FAANG •
Explore all Two Pointers problems & invariants →
How to Recognize This Pattern
- Common Strings interview pattern
- Optimal time: O(n^2)
Complexity Analysis
Time Complexity
O(n^2)
Space Complexity
O(1)
Tags
String
Frequently Asked Questions
What is the optimal time and space complexity for Minimum Number of Moves to Make Palindrome?
The optimal solution for Minimum Number of Moves to Make Palindrome achieves O(n^2) runtime complexity and requires O(1) auxiliary space. Time Complexity is O(n^2) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(1) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Minimum Number of Moves to Make Palindrome requires the Two Pointers pattern?
In technical interviews, look for key problem signals: Common Strings interview pattern; Optimal time: O(n^2). Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Two Pointers strategy.
What critical edge cases must be handled in Minimum Number of Moves to Make Palindrome?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Minimum Number of Moves to Make Palindrome?
Minimum Number of Moves to Make Palindrome is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Microsoft.