Maximize the Minimum Powered City

This is the optimal approach, building upon the binary search framework. The key improvement is optimizing the check(x) function to run in linear time, O(n). This is achieved by avoiding the repeated power calculations. We first pre-calculate the initial power of every city. Then, as we iterate through the cities and add new stations, we use a difference array (or a similar sliding window technique) to efficiently track the additional power (boost) these new stations provide to subsequent cities.

1. Define a search range for the minimum power, from 0 to sum(stations) + k.
2. Perform binary search on this range. For each mid value:

• Call an efficient check(mid) function.
• The check(targetPower) function works as follows:
  1. In O(n), pre-calculate initialPower[i] for all i using a prefix sum array on stations.
  2. Initialize a difference array boostDiff of size n+1 to all zeros.
  3. Initialize stationsLeft = k and currentBoost = 0.
  4. Iterate through each city i from 0 to n-1: a. Update the power from new stations: currentBoost += boostDiff[i]. b. Calculate total power: totalPower = initialPower[i] + currentBoost. c. If totalPower < targetPower, calculate needed stations. d. If needed > stationsLeft, return false. e. Decrement stationsLeft by needed. f. Increase currentBoost by needed as these new stations help city i immediately. g. Schedule the boost to be removed after its effect ends. The stations added for city i cover up to city i+2r. So, boostDiff[i+2r+1] -= needed (with bounds check).
  5. If the loop completes, return true.

3. If check(mid) is true, try for a higher power: ans = mid, low = mid + 1.
4. If check(mid) is false, we need a lower power: high = mid - 1.
5. Return the final ans.

The check(targetPower) function is optimized as follows:

1. Pre-computation: Calculate the initial power of every city based on the original stations array. This can be done in O(n) time using a prefix sum array. initialPower[i] = sum(stations[j]) for j in [i-r, i+r].
2. Greedy Iteration with Efficient Updates: We iterate through cities from i = 0 to n-1. We maintain a variable currentBoost which represents the extra power city i receives from all the new stations added for previous cities (j < i).
3. When we add needed stations for city i, they are placed at i+r and provide power to cities in the range [i, i+2r]. This means the currentBoost increases by needed starting from city i. This boost effect stops after city i+2r. We can model this efficiently using a difference array, boostDiff. When we add needed stations for city i, we've already added needed to currentBoost. To stop this effect later, we schedule a subtraction: boostDiff[i + 2*r + 1] -= needed.
4. At each city i, we update currentBoost by adding boostDiff[i]. The total power is then initialPower[i] + currentBoost. If this is less than targetPower, we add new stations and update our data structures as described.

Here is the efficient check function implementation:

private boolean check(long targetPower, int[] stations, int r, long k) {
    int n = stations.length;
    long[] prefixSum = new long[n + 1];
    for (int i = 0; i < n; i++) {
        prefixSum[i + 1] = prefixSum[i] + stations[i];
    }

    long[] initialPower = new long[n];
    for (int i = 0; i < n; i++) {
        int left = Math.max(0, i - r);
        int right = Math.min(n - 1, i + r);
        initialPower[i] = prefixSum[right + 1] - prefixSum[left];
    }

    long stationsLeft = k;
    long[] boostDiff = new long[n + 1];
    long currentBoost = 0;

    for (int i = 0; i < n; i++) {
        currentBoost += boostDiff[i];
        long currentTotalPower = initialPower[i] + currentBoost;

        if (currentTotalPower < targetPower) {
            long needed = targetPower - currentTotalPower;
            if (needed > stationsLeft) {
                return false;
            }
            stationsLeft -= needed;
            currentBoost += needed;
            int endEffectIndex = i + 2 * r + 1;
            if (endEffectIndex < n) {
                boostDiff[endEffectIndex] -= needed;
            }
        }
    }
    return true;
}