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Algorithms/Dynamic Programming/lis_length.cpp

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+// Length of Longest Increasing Subsequence
+//n(logn) time complexity
+#include <iostream>
+#include <vector>
+
+// Binary search (note boundaries in the caller)
+int CeilIndex(std::vector<int>& v, int l, int r, int key)
+{
+    while (r - l > 1) {
+        int m = l + (r - l) / 2;
+        if (v[m] >= key)
+            r = m;
+        else
+            l = m;
+    }
+
+    return r;
+}
+
+int LongestIncreasingSubsequenceLength(std::vector<int>& v)
+{
+    if (v.size() == 0)
+        return 0;
+
+    std::vector<int> tail(v.size(), 0);
+    int length = 1; // always points empty slot in tail
+
+    tail[0] = v[0];
+    for (size_t i = 1; i < v.size(); i++) {
+
+        // new smallest value
+        if (v[i] < tail[0])
+            tail[0] = v[i];
+
+        // v[i] extends largest subsequence
+        else if (v[i] > tail[length - 1])
+            tail[length++] = v[i];
+
+        // v[i] will become end candidate of an existing
+        // subsequence or Throw away larger elements in all
+        // LIS, to make room for upcoming grater elements
+        // than v[i] (and also, v[i] would have already
+        // appeared in one of LIS, identify the location
+        // and replace it)
+        else
+            tail[CeilIndex(tail, -1, length - 1, v[i])] = v[i];
+    }
+
+    return length;
+}
+
+int main()
+{
+    std::vector<int> v;
+    v.push_back(2);v.push_back(5);v.push_back(3);v.push_back(7);v.push_back(11);v.push_back(8);v.push_back(10);v.push_back(13);v.push_back(6);
+    std::cout << "Length of Longest Increasing Subsequence is "
+              << LongestIncreasingSubsequenceLength(v) << '\n';
+    return 0;
+}