The Palindrome Removal problem asks for the minimum number of moves to remove all elements from an array, where each move removes a palindromic subarray. This hard coding problem uses interval DP, where dp[i][j] represents the minimum moves to remove arr[i..j]. The array and dynamic programming interview pattern is the core.
Microsoft asks this hard interval DP problem because it tests complex state transitions: when removing palindromes, adjacent elements can merge or form new palindromes. The key insight is that equal elements at the endpoints can share their removal. The array and dynamic programming interview pattern is fully exercised.
Interval DP. dp[i][j] = minimum moves to remove arr[i..j]. Base: dp[i][i]=1. Transition: split at any k in [i,j-1]: dp[i][j] = min(dp[i][k] + dp[k+1][j]). Optimization: if arr[i] == arr[j], dp[i][j] = dp[i+1][j-1] if j-i≤2, or try: dp[i][j] = dp[i][k] + dp[k+1][j-1] when arr[k] == arr[j]. Also: dp[i][j] = 1 if arr[i..j] itself is a palindrome.
arr=[1,3,4,1,5]. dp[0][0]=1,dp[1][1]=1,...dp[3][3]=1. dp[0][3]: arr[0]==arr[3]=1. Try dp[1][2]+dp stuff. dp[0][3] = dp[1][2] = 2 (remove "34" in 2 moves). So remove "1341" as... "1_1" wrapping around "34" → dp[0][3]=2+1=? Actually dp[1][2]=2, and since arr[0]==arr[3], dp[0][3]=dp[1][2]=2. dp[0][4]=min(dp[0][3]+dp[4][4])=3, or other splits. Answer = 3.
Palindrome Removal is an advanced interval DP problem. The critical transition — when endpoints are equal, they can be removed together with the interior — is non-trivial. Practice standard interval DP (Burst Balloons, Strange Printer, Stone Merging) before attempting this. The key debugging technique: verify small cases manually before coding the full DP.
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