Earliest Second to Mark Indices II

• Iterate through possible seconds s from n to m.
• For each s, check if it's possible to mark all indices using a helper function canMarkAll(s).
• The canMarkAll(s) function implements a greedy strategy:
  • Calculate the baseline operations needed: n marks + sum(nums) decrements.
  • Determine which indices can be reset to 0 within s seconds.
  • For each such index i, if nums[i] > 1, using the 'set' operation saves nums[i] - 1 operations. Greedily take all such savings.
  • Sum up the total savings and reduce the opsNeeded.
  • If s is greater than or equal to the final opsNeeded, it's possible.
• Return the first s for which canMarkAll(s) is true.

The core of this approach is a helper function, canMarkAll(s), which determines if it's possible to mark all n indices within s seconds. We can iterate s from n to m. The first s for which canMarkAll(s) returns true is our answer. If the loop finishes without finding such an s, it's impossible.

canMarkAll(s) Logic:

The strategy within canMarkAll(s) is based on a greedy choice. For a given time s, we have s seconds to perform operations. The most powerful operation is setting nums[i] to 0, which saves nums[i] decrement operations at the cost of one second. This is a net saving of nums[i] - 1 operations. This saving is beneficial only if nums[i] > 1.

1. First, calculate the total number of operations required if we only use decrements and marks. This would be n (for marks) + sum(nums).
2. For the given s seconds, identify all indices that can be set to 0. An index i can be set to 0 if it appears in changeIndices at least once within the first s seconds. To make the best use of this operation, we should use it at the latest possible second for each such index, but the greedy choice doesn't depend on the timing, just on its availability.
3. For each index i that can be set to 0, we greedily decide whether to use the 'set' operation. We use it if it reduces the total number of operations, which happens when nums[i] - 1 > 0, or nums[i] > 1.
4. We calculate the total savings from these greedy choices and subtract it from the initial total operations needed.
5. If the final number of operations needed is less than or equal to s, then it's possible. Otherwise, it's not.

Algorithm:

1. Iterate s from n to m.
2. For each s, call canMarkAll(s): a. Calculate totalSum = sum(nums[i]) for all i. b. Calculate opsNeeded = n + totalSum. c. Find the last occurrence last[i] for each index i within changeIndices[1...s]. d. For each index i that has a last[i] > 0: i. If nums[i-1] > 1, we choose to use the 'set' operation. The saving is nums[i-1] - 1. ii. Subtract this saving from opsNeeded. e. If s >= opsNeeded, then it's possible. Return s as the answer.
3. If the loop completes, it's impossible. Return -1.

1class Solution {2    public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {3        int n = nums.length;4        int m = changeIndices.length;5 6        for (int s = n; s <= m; s++) {7            if (canMarkAll(s, nums, changeIndices)) {8                return s;9            }10        }11        return -1;12    }13 14    private boolean canMarkAll(int s, int[] nums, int[] changeIndices) {15        int n = nums.length;16        int[] last = new int[n + 1];17        for (int i = 0; i < s; i++) {18            last[changeIndices[i]] = i + 1;19        }20 21        long totalSum = 0;22        for (int num : nums) {23            totalSum += num;24        }25 26        long opsNeeded = (long)n + totalSum;27        int settableCount = 0;28        for(int i=1; i<=n; ++i) {29            if(last[i] > 0) {30                settableCount++;31            }32        }33        if(settableCount < n) { // Not all indices appear34            // This check is not strictly necessary as the main logic handles it,35            // but it highlights that some indices must be decremented.36        }37 38        for (int i = 1; i <= n; i++) {39            if (last[i] > 0) { // If index i is settable40                if (nums[i - 1] > 1) {41                    opsNeeded -= (nums[i - 1] - 1);42                }43            }44        }45 46        return s >= opsNeeded;47    }48}

• The problem has a monotonic property: if all indices can be marked in s seconds, they can also be marked in s+1 seconds.
• This allows for binary searching on the answer s in the range [1, m].
• For each mid value of s in the binary search, use the same greedy canMarkAll(s) function from the previous approach.
• If canMarkAll(mid) is true, we try to find an even earlier second by searching in the lower half [low, mid-1].
• If canMarkAll(mid) is false, we need more seconds, so we search in the upper half [mid+1, high].

We can binary search for the minimum s in the range [1, m]. For each mid value in the binary search, we use the same canMarkAll(mid) helper function as in the previous approach to check if it's a feasible number of seconds.

• If canMarkAll(mid) is true, it means mid could be our answer, but there might be an even smaller s that works. So, we try smaller values by setting high = mid - 1 and storing mid as a potential answer.
• If canMarkAll(mid) is false, mid seconds are not enough. We need more time, so we set low = mid + 1.

The canMarkAll(s) function remains the same, employing the greedy strategy of minimizing total operations.

Algorithm:

1. Initialize low = 1, high = m, and ans = -1.
2. While low <= high: a. mid = low + (high - low) / 2. b. Call canMarkAll(mid). c. If canMarkAll(mid) is true: i. mid is a potential answer. Store it: ans = mid. ii. Try for a smaller s: high = mid - 1. d. Else (canMarkAll(mid) is false): i. Need more time: low = mid + 1.
3. Return ans.

1class Solution {2    public int earliestSecondToMarkIndices(int[] nums, int[] changeIndices) {3        int n = nums.length;4        int m = changeIndices.length;5        int ans = -1;6        int low = 1, high = m;7 8        while (low <= high) {9            int mid = low + (high - low) / 2;10            if (canMarkAll(mid, nums, changeIndices)) {11                ans = mid;12                high = mid - 1;13            } else {14                low = mid + 1;15            }16        }17        return ans;18    }19 20    private boolean canMarkAll(int s, int[] nums, int[] changeIndices) {21        int n = nums.length;22        // A quick check: we need at least n seconds to mark n indices.23        if (s < n) {24            return false;25        }26 27        int[] last = new int[n + 1];28        for (int i = 0; i < s; i++) {29            last[changeIndices[i]] = i + 1;30        }31 32        // Check if all indices appear at least once if their value > 033        // This is a necessary condition for the greedy strategy to work34        // as we need a 'set' operation for large nums values.35        // However, the main logic below implicitly handles this.36        int settableCount = 0;37        for(int i=1; i<=n; ++i) {38            if(last[i] > 0) {39                settableCount++;40            }41        }42        // If an index has a large value and never appears in changeIndices, it might be impossible.43        // The total ops logic handles this correctly.44 45        long totalSum = 0;46        for (int num : nums) {47            totalSum += num;48        }49 50        long opsNeeded = (long)n + totalSum;51 52        for (int i = 1; i <= n; i++) {53            if (last[i] > 0) { // If index i is settable54                // The 'set' operation costs 1 second and saves nums[i-1] decrements.55                // Net saving is nums[i-1] - 1. We do this if saving is positive.56                if (nums[i - 1] > 1) {57                    opsNeeded -= (nums[i - 1] - 1);58                }59            }60        }61 62        return s >= opsNeeded;63    }64}