![]() ![]() With "use_manacher=false" so that it falls back to a straightforward O(n 2) algorithm, where we start from each possible center point, and expand to both sides until we reach the length of the string we have: Total CPU time : 85.582152 sec "Solution is correct" means the output matches what's provided by HackerRank. Here are the results with "use_manacher=true" and HackerRank input file input16.txt, which is a complex test case with almost 10 5 characters. ![]() (String.format(" Length calculation took %.6f sec", (float)(tEndNS - tOddNS) / 1.0e9)) (String.format(" Odd palindromes took %.6f sec", (float)(tOddNS - tEvenNS) / 1.0e9)) (String.format(" Even palindromes took %.6f sec", (float)(tEvenNS - tStartNS) / 1.0e9)) (String.format("Total CPU time : %.6f sec", (float)(tEndNS - tStartNS) / 1.0e9)) (String.format("Mismatch to solution: lengths = %d (should be %d)", i, lengths, solution)) Solution = Integer.valueOf(solin.readLine()) ("Reading solution file " + solutionFile) īufferedReader solin = new BufferedReader(new InputStreamReader(new FileInputStream(solfile))) Long tEndNS = threadMXBean.getCurrentThreadCpuTime() Int len = Math.min(plenOdd, Math.min(spaceLeft*2+1, spaceRight*2+1)) Int spaceRight = (i >= k) ? (N + k - i - 1) : (k - i - 1) Len = Math.min(plenEven, Math.min(spaceLeft*2, spaceRight*2)) Int spaceRight = (i > k) ? (N + k - i) : (k - i) Part 3 : Find maximum palindrome for Skįor(int k=0 k= k) ? (i - k) : (N + i - k) Long tOddNS = threadMXBean.getCurrentThreadCpuTime() Int plen_best = dist_to_best_right * 2 + 1 Int i2 = (best_left + dist_to_best_right) % N Int dist_to_best_right = (best_right - i + N) % N Long tEvenNS = threadMXBean.getCurrentThreadCpuTime() īest_right = 0 best_left = 0 best_plen = 1 Int i2 = (best_left + best_right - i + 1 + N) % N Int best_right = N-1, best_left = 0, best_plen = 0 Long tStartNS = threadMXBean.getCurrentThreadCpuTime() ThreadMXBean threadMXBean = ManagementFactory.getThreadMXBean() ("Reading file " + inputFile) įileInputStream fis = new FileInputStream(file) īufferedReader in = new BufferedReader(new InputStreamReader(fis)) String solutionFile = "D:\\Home\\Java\\AlgorithmPractice\\output16.txt" String inputFile = "D:\\Home\\Java\\AlgorithmPractice\\input16.txt" I have written the following Java code, which is a modification of the Manacher's algorithm in a rotated context: package It also adds another layer of complexity by asking us to find the lengths for each rotate string.Ī similar question has been posted here, but I couldn't extract enough information to get my code working fast enough. In this hackerRank challenge, the bigger string has a length limit of 10 5. ![]() ![]() The traditional palindrome problem is basically to find the length of longest symmetric substrings (palindromes) within a bigger string. I have been trying to solve the problem of circular palindrome all day, as part of a HackerRank challenge. ![]()
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