Minimum Cost to Make All Characters Equal

• Initialize totalCost = 0.
• Iterate through the string from the second character (i = 1 to n-1).
• At each position i, compare s.charAt(i) with s.charAt(i-1).
• If they are different, it means there is a mismatch that must be resolved.
• The cost to resolve this mismatch is the minimum of flipping the prefix s[0...i-1] (cost i) or flipping the suffix s[i...n-1] (cost n-i).
• Add min(i, n - i) to totalCost.
• After the loop finishes, totalCost will hold the total minimum cost.

This approach is built on the same fundamental insight as the DP solution: the total minimum cost is the sum of the minimum costs to resolve each individual mismatch between adjacent characters. A mismatch occurs at index i-1 if s[i-1] != s[i]. To resolve this, we must apply an operation that flips one character but not the other. The two options are:

1. A prefix flip up to i-1, with cost (i-1) + 1 = i.
2. A suffix flip from i, with cost n - i.

The minimum cost to resolve this specific mismatch is min(i, n - i). Since these choices are independent for each mismatch, we can simply iterate through the string, find all mismatches, calculate the minimum cost for each, and sum them up. We use a single variable, totalCost, to keep a running sum, which optimizes space.

1class Solution {2    public long minimumCost(String s) {3        int n = s.length();4        long totalCost = 0;5 6        // Iterate through the string to find adjacent characters that are different.7        for (int i = 1; i < n; i++) {8            if (s.charAt(i) != s.charAt(i-1)) {9                // A mismatch at index i-1 needs to be fixed.10                // We can either flip the prefix ending at i-1 (cost i)11                // or flip the suffix starting at i (cost n-i).12                // We choose the cheaper of the two operations.13                totalCost += Math.min(i, n - i);14            }15        }16 17        return totalCost;18    }19}

• The key insight is that the total cost is the sum of costs to fix each position i where s[i] != s[i+1].
• For each such "mismatch", we must perform an operation that flips one of the characters but not the other. The two options are a prefix flip up to i (cost i+1) or a suffix flip from i+1 (cost n-(i+1)).
• We can use dynamic programming. Let dp[i] be the minimum cost to make the prefix s[0...i] uniform.
• The recurrence relation is: dp[i] = dp[i-1] if s[i] == s[i-1], and dp[i] = dp[i-1] + min(i, n-i) if s[i] != s[i-1].
• The base case is dp[0] = 0.
• The final answer is dp[n-1].

The core idea is that to make the string uniform, we must eliminate all indices i where s[i] != s[i+1]. To resolve such a mismatch, we must apply an operation that splits i and i+1. The two choices are a prefix flip up to i (cost i+1) or a suffix flip from i+1 (cost n-(i+1)). Since fixing one mismatch doesn't create or resolve another, we can consider each mismatch independently and choose the cheaper operation for each.

We can define a DP state dp[i] as the minimum cost to resolve all mismatches in the prefix s[0...i]. We build the solution from left to right.

1class Solution {2    public long minimumCost(String s) {3        int n = s.length();4        if (n <= 1) {5            return 0;6        }7 8        // dp[i] stores the minimum cost to make the prefix s[0...i] uniform.9        long[] dp = new long[n];10        dp[0] = 0; // A single character prefix is always uniform.11 12        for (int i = 1; i < n; i++) {13            // Start with the cost to make the previous prefix uniform.14            dp[i] = dp[i-1];15            16            // If there is a mismatch between s[i] and s[i-1], we must fix it.17            if (s.charAt(i) != s.charAt(i-1)) {18                // To fix the mismatch at index i-1, we need an operation that19                // splits i-1 and i.20                // Option 1: Prefix flip up to i-1. Cost = (i-1) + 1 = i.21                // Option 2: Suffix flip from i. Cost = n - i.22                long costToFix = Math.min(i, n - i);23                dp[i] += costToFix;24            }25        }26 27        return dp[n - 1];28    }29}

• Model the problem as a shortest path problem on a graph.
• Each unique binary string of length n is a node in the graph.
• An edge exists between two string nodes if one can be transformed into the other by a single operation. The weight of the edge is the cost of that operation.
• Use Dijkstra's algorithm to find the shortest path from the initial string s to any uniform string (all '0's or all '1's).
• Use a priority queue to explore states with the lowest cost first and a hash map to store minimum costs to visited states to avoid cycles.

The state in our search is the current string, and the cost is the accumulated cost of operations. We use a priority queue to always explore the state with the minimum accumulated cost first, which is the core of Dijkstra's algorithm. A map or hash table is used to keep track of the minimum cost found so far for each string to avoid cycles and redundant computations.

The algorithm starts with the initial string s and a cost of 0. In each step, it extracts the string with the lowest cost from the priority queue. If this string is uniform, we have found the minimum cost. Otherwise, it generates all possible next strings by applying each of the 2n operations (prefix flips for i=0..n-1 and suffix flips for i=0..n-1). For each new string, if a cheaper path to it is found, its cost is updated, and it's added to the priority queue.

1// This approach is purely theoretical for the given constraints and would time out.2// A full implementation is impractical due to the massive state space.3class Solution {4    // Conceptual structure of a brute-force Dijkstra's approach.5    public long minimumCost(String s) {6        int n = s.length();7        PriorityQueue<Pair<Long, String>> pq = new PriorityQueue<>(Comparator.comparingLong(Pair::getKey));8        Map<String, Long> minCosts = new HashMap<>();9 10        pq.add(new Pair<>(0L, s));11        minCosts.put(s, 0L);12 13        while (!pq.isEmpty()) {14            Pair<Long, String> current = pq.poll();15            long cost = current.getKey();16            String currentS = current.getValue();17 18            if (isUniform(currentS)) {19                return cost;20            }21 22            if (cost > minCosts.get(currentS)) {23                continue;24            }25 26            // Explore all 2n possible operations27            for (int i = 0; i < n; i++) {28                // ... logic to apply prefix and suffix flips ...29                // ... update pq and minCosts if a cheaper path is found ...30            }31        }32        return -1; // Should not be reached33    }34 35    private boolean isUniform(String s) {36        if (s.length() <= 1) return true;37        char first = s.charAt(0);38        for (int i = 1; i < s.length(); i++) {39            if (s.charAt(i) != first) return false;40        }41        return true;42    }43}