Minimum Insertions to Make String Palindrome
A hard Dynamic Programming problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
- Platform
- LeetCode
The problem
Given a string, find the minimum number of insertions needed to make it a palindrome. You can insert characters at any position.
Example 1
- Input
- s = "zzazz"
- Output
- 0
- Why
- "zzazz" is already a palindrome, no insertions needed.
Example 2
- Input
- s = "mbadm"
- Output
- 2
- Why
- Insert two characters to make it "mbdadbm" or similar palindrome.
Example 3
- Input
- s = "leetcode"
- Output
- 5
- Why
- 5 insertions are needed to make "leetcode" a palindrome.
Constraints
- 1 <= s.length <= 500
- s consists of lowercase English letters
How to think about it
Updated 2026-09-09Every character already forming part of the longest palindromic subsequence is already matched and balanced. The minimum insertions needed to balance the rest is the total length of the string minus the length of its longest palindromic subsequence.
Approaches, worst first
Recursive boundary matching
time O(2^n) · space O(n)
Compare characters at left and right pointers. If they match, advance both. If they differ, branch into inserting at left or right: 1 + min(recur(i+1, j), recur(i, j-1)). Explores exponential states.
Longest palindromic subsequence reduction
time O(n^2) · space O(n^2)
Reverse s to form s_rev. Compute LCS between s and s_rev using 2D dynamic programming. The answer is n - lcs_length.
Interval DP with rolling 1D arrayWrite this one
time O(n^2) · space O(n)
Let dp[j] be min insertions for substring s[i..j]. Iterate i from n - 1 down to 0, updating dp[j] using character matching or 1 + min(dp[j], dp[j-1]).
Where people lose marks · 3
- Forgetting that a string of length 1 is already a palindrome requiring 0 insertions.
- Attempting greedy insertion from the outside inward without backtracking on ties.
- Off-by-one errors when reversing indices or mapping substrings to DP table cells.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
Track this in your role's order
Pick your target role and all 370 problems — including this one — resequence to what that interview actually asks. Free.
Start freeMore Dynamic Programming problems
- Longest Palindromic SubsequenceMedium
- Minimum Insertions/Deletions to Convert StringMedium
- Shortest Common SupersequenceHard
- Unbounded KnapsackMedium
- Maximum Rectangle in Binary MatrixHard
- Longest Arithmetic SubsequenceMedium
- Boolean ParenthesizationHard
- Number of Ways to Reach Destination (DP on Grid)Medium
Problem set and role mapping as of .