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This approach involves treating each tap as an interval [i-ranges[i], i+ranges[i]], and seeking the minimal number of such intervals required to cover the entire garden range [0, n]. The key is to employ a greedy algorithm which iteratively selects the interval that extends the coverage furthest at each point. Specifically, from each starting point, choose the interval that reaches farthest within the current range, then jump to the furthest point reached by the selected intervals and repeat the process.
Time Complexity: O(n), as the algorithm processes the tap ranges in a single pass.
Space Complexity: O(n), due to the storage for the maxRange array.
1#include <stdio.h>
2#include <string.h>
3#include <limits.h>
4
5int minTaps(int n, int* ranges,
This C solution first transforms the ranges into maximum right reach for each start position. It employs a greedy approach, moving through the garden while extending the reach of water as far as possible with the least number of taps activated. If a gap or unwatered region is found, it returns -1.
This approach takes on the problem by formulating it into a dynamic programming challenge. Here, the idea is to maintain an array dp[], where dp[i] denotes the minimum number of taps required to water up to position i. Start by initializing dp[i] with infinity, except dp[0]=0, then update each dp[i] by considering if current tap can extend water coverage from any previous dp positions.
Time Complexity: O(n^2), due to the potential iteration across all tap ranges for each garden position.
Space Complexity: O(n), as a dp array of size n+1 is maintained.
In Java, maintaining a dp array initialized with maximum values and updating it using available tap ranges solves the problem by ensuring the smallest tap set is used to water the entire garden. The solution checks for the condition where no coverage is possible.