Zero Array Transformation IV

• For each possible number of queries k from 1 to queries.length:
  • Call a function canMakeZero(k) to check if nums can be turned into a zero array.
  • If canMakeZero(k) returns true, then k is the minimum number of queries, so return k.
• If the loop completes without finding a solution, return -1.
• The canMakeZero(k) function checks each index j of nums:
  • It gathers all val_i from the first k queries that are applicable to index j.
  • It uses a recursive backtracking function to check if nums[j] can be formed by a sum of a subset of these values.
  • If any index j fails this check, canMakeZero(k) returns false.
  • If all indices pass, it returns true.

The core idea is to test each potential answer k sequentially. For a given k, we must determine if there's a way to apply the first k queries to make every nums[j] zero. This decomposes into n independent subset sum problems. For each nums[j], we need to find if it can be expressed as a sum of some of the val_i from the first k queries that cover index j.

The backtracking function canFormSum explores all subsets of the applicable query values to see if any sum up to the target nums[j]. It does this by making two recursive calls at each step: one where the current value is included in the sum, and one where it's excluded.

1class Solution {2    public int zeroArray(int[] nums, int[][] queries) {3        for (int k = 1; k <= queries.length; k++) {4            if (canMakeZero(nums, queries, k)) {5                return k;6            }7        }8        return -1;9    }10 11    private boolean canMakeZero(int[] nums, int[][] queries, int k) {12        for (int j = 0; j < nums.length; j++) {13            if (nums[j] == 0) continue;14            java.util.List<Integer> options = new java.util.ArrayList<>();15            for (int i = 0; i < k; i++) {16                if (queries[i][0] <= j && j <= queries[i][1]) {17                    options.add(queries[i][2]);18                }19            }20            if (!canFormSum(nums[j], options, 0)) {21                return false;22            }23        }24        return true;25    }26 27    private boolean canFormSum(int target, java.util.List<Integer> options, int index) {28        if (target == 0) return true;29        if (target < 0 || index == options.size()) return false;30        31        // Exclude current option32        if (canFormSum(target, options, index + 1)) {33            return true;34        }35        // Include current option36        if (canFormSum(target - options.get(index), options, index + 1)) {37            return true;38        }39        return false;40    }41}

• The problem has a monotonic property: if the array can be zeroed with k queries, it can also be zeroed with k+1 queries. This allows for binary search on the answer k.
• Binary search for k in the range [0, queries.length].
• For each mid value in the binary search, call a helper function canMakeZero(mid).
• The canMakeZero(k) function checks if it's possible to zero the array using the first k queries.
  • For each index j, it solves the subset sum problem for the target nums[j] using values from the first k applicable queries.
  • This subset sum problem is solved efficiently using dynamic programming.
  • A boolean array dp of size nums[j] + 1 is used, where dp[s] is true if sum s is achievable.
• If canMakeZero(mid) is true, we might find a better answer with fewer queries, so we search in the lower half (high = mid - 1). Otherwise, we need more queries and search in the upper half (low = mid + 1).

Instead of a linear scan for k, we can significantly speed up the search by using binary search. The search space for k is from 0 to queries.length.

For a given k (let's say mid in the binary search), the canMakeZero(k) function determines feasibility. For each index j, we need to check if nums[j] can be formed by summing a subset of values from the first k queries that apply to j. This is a classic subset sum problem, which we solve with dynamic programming to avoid the exponential complexity of backtracking.

We create a boolean DP array dp of size nums[j] + 1. dp[s] is true if a total decrement of s is possible. We initialize dp[0] = true. Then, for each applicable query value v, we update the dp table: dp[s] = dp[s] || dp[s - v] for s from nums[j] down to v. If dp[nums[j]] is false after considering all applicable queries, then k is not sufficient.

1class Solution {2    public int zeroArray(int[] nums, int[][] queries) {3        int low = 0, high = queries.length;4        int ans = -1;5 6        while (low <= high) {7            int mid = low + (high - low) / 2;8            if (canMakeZero(nums, queries, mid)) {9                ans = mid;10                high = mid - 1;11            } else {12                low = mid + 1;13            }14        }15        return ans;16    }17 18    private boolean canMakeZero(int[] nums, int[][] queries, int k) {19        if (k == 0) {20            for (int num : nums) {21                if (num != 0) return false;22            }23            return true;24        }25        26        for (int j = 0; j < nums.length; j++) {27            if (nums[j] == 0) continue;28            29            boolean[] dp = new boolean[nums[j] + 1];30            dp[0] = true;31 32            for (int i = 0; i < k; i++) {33                if (queries[i][0] <= j && j <= queries[i][1]) {34                    int val = queries[i][2];35                    for (int s = nums[j]; s >= val; s--) {36                        dp[s] = dp[s] || dp[s - val];37                    }38                }39            }40            if (!dp[nums[j]]) {41                return false;42            }43        }44        return true;45    }46}

• Initialize a 2D boolean DP table possible[n][max_val + 1], where max_val is the maximum value in nums.
• possible[j][s] will store whether a total decrement of s is achievable for index j.
• For each index j, initialize possible[j][0] = true.
• Iterate through the queries from i = 0 to queries.length - 1.
  • For the current query [l, r, val], update the possible table for all affected indices j from l to r.
  • The update rule is: possible[j][s] = possible[j][s] || possible[j][s - val] for s from max_val down to val.
• After processing each query i, check if a solution has been found:
  • Assume a solution is possible (all_zeroable = true).
  • For each index j, check if possible[j][nums[j]] is true. If not, set all_zeroable = false and break.
  • If all_zeroable is still true, we have found the minimum k. Return i + 1.
• If the loop finishes, no solution was found. Return -1.

This approach builds the solution incrementally. We maintain a DP table, possible[j][s], which tells us if a sum s can be formed for index j using the queries processed so far. We iterate through the queries one by one. For each query, we update the possible sums for all indices it affects. After each query i is processed, we check if it's now possible to zero out the entire array. This is done by checking if for every index j, possible[j][nums[j]] is true. If this condition holds, we have found our answer, and it must be the minimum k because we are processing queries in order. The minimum number of queries is therefore i + 1.

This method avoids the overhead of binary search and re-computation within the check function, leading to a more direct and often faster solution.

1class Solution {2    public int zeroArray(int[] nums, int[][] queries) {3        int n = nums.length;4        int maxVal = 0;5        for (int num : nums) {6            maxVal = Math.max(maxVal, num);7        }8 9        if (maxVal == 0) {10            return 0;11        }12 13        boolean[][] possible = new boolean[n][maxVal + 1];14        for (int j = 0; j < n; j++) {15            possible[j][0] = true;16        }17 18        for (int i = 0; i < queries.length; i++) {19            int l = queries[i][0];20            int r = queries[i][1];21            int val = queries[i][2];22 23            for (int j = l; j <= r; j++) {24                for (int s = maxVal; s >= val; s--) {25                    possible[j][s] = possible[j][s] || possible[j][s - val];26                }27            }28 29            boolean allZeroable = true;30            for (int j = 0; j < n; j++) {31                if (!possible[j][nums[j]]) {32                    allZeroable = false;33                    break;34                }35            }36 37            if (allZeroable) {38                return i + 1;39            }40        }41 42        return -1;43    }44}