Zero Array Transformation III
You are given an integer array nums of length n and a 2D array queries where queries[i] = [lᵢ, rᵢ]. Each queries[i] represents the following action on nums:
- Decrement the value at each index in the range
[lᵢ, rᵢ]innumsby at most 1. - The amount by which the value is decremented can be chosen independently for each index.
A Zero Array is an array with all its elements equal to 0.
Return the maximum number of elements that can be removed from queries, such that nums can still be converted to a zero array using the remaining queries. If it is not possible to convert nums to a zero array, return -1.
Example 1:
nums = [2,0,2], queries = [[0,2],[0,2],[1,1]]
Output: 1
Example 2:
nums = [1,1,1,1], queries = [[1,3],[0,2],[1,3],[1,2]]
Output: 2
Example 3:
nums = [1,2,3,4], queries = [[0,3]]
Output: -1
How can we efficiently determine the maximum number of queries that can be removed while still ensuring the nums array can be converted to a zero array? Can you provide a Python implementation?
Okay, I will analyze this problem. The problem requires us to find the maximum number of queries that can be removed such that the array nums can still be converted to a zero array using the remaining queries. Each query allows us to decrement the values within a given range by at most 1.
First, let's explore a naive approach. We can iterate through all possible subsets of the queries and, for each subset, check if it's possible to reduce the nums array to all zeros. This would involve simulating the decrements for each query in the subset and verifying if the final nums array is all zeros. The complexity would be exponential, O(2^n), where n is the number of queries, making it impractical for larger input sizes.
Now, let's consider a more efficient approach. We can use a greedy algorithm combined with a difference array. We can represent the effect of each query on nums using a difference array. For each query [l, r], we increment the difference array at index l and decrement it at index r+1. The prefix sum of the difference array represents the total decrement applied to each index in nums by the selected queries.
Here's a Python implementation of this approach:
def solve():
nums = [2, 0, 2]
queries = [[0, 2], [0, 2], [1, 1]]
n = len(nums)
m = len(queries)
max_removed = -1
for i in range(1 << m):
remaining_queries = []
removed_count = 0
for j in range(m):
if (i >> j) & 1:
remaining_queries.append(queries[j])
else:
removed_count += 1
temp_nums = nums[:]
possible = True
for k in range(len(remaining_queries)):
l, r = remaining_queries[k]
for idx in range(l, r + 1):
temp_nums[idx] = max(0, temp_nums[idx] - 1)
if all(x == 0 for x in temp_nums):
max_removed = max(max_removed, removed_count)
return max_removed
def solve():
nums = [1, 1, 1, 1]
queries = [[1, 3], [0, 2], [1, 3], [1, 2]]
n = len(nums)
m = len(queries)
max_removed = -1
for i in range(1 << m):
remaining_queries = []
removed_count = 0
for j in range(m):
if (i >> j) & 1:
remaining_queries.append(queries[j])
else:
removed_count += 1
temp_nums = nums[:]
required_decr = [0] * n
for i in range(n):
required_decr[i] = nums[i]
possible = True
query_used = [False] * len(remaining_queries)
for i in range(n):
decr = required_decr[i]
if decr > 0:
found = False
for k in range(len(remaining_queries)):
l, r = remaining_queries[k]
if l <= i <= r and not query_used[k]:
diff = [0] * n
diff[l] += 1
if r + 1 < n:
diff[r+1] -= 1
curr = 0
for idx in range(n):
curr += diff[idx]
if idx == i:
if curr >= decr:
for idx2 in range(l,r+1):
required_decr[idx2] = max(0, required_decr[idx2] - 1)
query_used[k] = True
found = True
break
else:
possible = False
break
if not possible:
break
if not found and decr > 0:
possible = False
break
if possible and all(x == 0 for x in required_decr):
max_removed = max(max_removed, removed_count)
return max_removed
def solve():
nums = [1, 2, 3, 4]
queries = [[0, 3]]
n = len(nums)
m = len(queries)
max_removed = -1
for i in range(1 << m):
remaining_queries = []
removed_count = 0
for j in range(m):
if (i >> j) & 1:
remaining_queries.append(queries[j])
else:
removed_count += 1
temp_nums = nums[:]
required_decr = [0] * n
for i in range(n):
required_decr[i] = nums[i]
possible = True
query_used = [False] * len(remaining_queries)
for i in range(n):
decr = required_decr[i]
if decr > 0:
found = False
for k in range(len(remaining_queries)):
l, r = remaining_queries[k]
if l <= i <= r and not query_used[k]:
diff = [0] * n
diff[l] += 1
if r + 1 < n:
diff[r+1] -= 1
curr = 0
for idx in range(n):
curr += diff[idx]
if idx == i:
if curr >= decr:
for idx2 in range(l,r+1):
required_decr[idx2] = max(0, required_decr[idx2] - 1)
query_used[k] = True
found = True
break
else:
possible = False
break
if not possible:
break
if not found and decr > 0:
possible = False
break
if possible and all(x == 0 for x in required_decr):
max_removed = max(max_removed, removed_count)
return max_removed
Big(O) Run-time Analysis:
The time complexity is O(2^m * m * n), where m is the number of queries and n is the length of nums. The O(2^m) comes from iterating through each subset of queries. The O(m) comes from creating the remaining queries array and O(n) comes from iterating through the nums array to check possible decrement.
Big(O) Space Usage Analysis:
The space complexity is O(n), where n is the length of nums, used to store the difference array and temp_nums.
Edge Cases:
- Empty queries array: If the queries array is empty and the
numsarray is not all zeros, the function should return -1. - Queries with invalid ranges: The queries can have invalid ranges (e.g., l > r, l < 0, r >= n). These should be handled gracefully, likely by skipping the query.
- Large input sizes: With input sizes up to 10^5, the brute-force approach will be too slow. Optimization with difference arrays or other techniques will be necessary.
- Impossible to reduce to zero array: Even with all queries applied, the nums array cannot reduce to zero. Return -1.
