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In this video, we solve LeetCode 1653: Minimum Deletions to Make String Balanced. 🔹 Problem Summary: You are given a string consisting only of 'a' and 'b'. A string is balanced if: There is NO pair (i, j) such that: i j AND s[i] = 'b' AND s[j] = 'a' Goal → Find minimum deletions to make the string balanced. ----------------------------------------------------- 🔹 Intuition Invalid Pattern → "ba" So we must either: Delete 'b' from left side OR Delete 'a' from right side ----------------------------------------------------- 🔹 Approach Explained We process string from left to right. Maintain: • countB → Number of 'b' seen so far • dp[i] → Minimum deletions till index i Transitions: If current char = 'a': Option 1 → Delete this 'a' → dp[i-1] + 1 Option 2 → Delete all previous 'b' → countB Take minimum. If current char = 'b': No conflict → Just carry previous dp value Increase countB. ----------------------------------------------------- 🔹 Why This Works We dynamically decide: Delete current character OR Delete conflicting previous characters. ----------------------------------------------------- 🔹 Complexity Time Complexity: O(n) Space Complexity: O(n) (Can be optimized to O(1)) ----------------------------------------------------- 🔹 Concepts Covered ✔ Dynamic Programming ✔ Prefix Counting ✔ Greedy Decision ✔ String Processing ✔ Optimization Thinking ----------------------------------------------------- Clean Linear Time DP problem — Very Important for Interviews. #LeetCode1653 #MinimumDeletionsStringBalanced #DPOnStrings #DynamicProgramming #GreedyDP #LeetCodeDaily #DSA #InterviewPreparation #FAANGPreparation #CompetitiveProgramming #StudyPlacement #StringDP #PrefixCount #CodingInterview #LeetCodeSolutions