Greedy-02 Solutions Posted

01-task-scheduler.cpp

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+/*
+TC - O(N) - performed linear scans only
+SC - O(26) i.e. O(1) - used extra space to hold constant freq counts for lowercase alphaeb
+Issues Faced - that idle slots logic does make sense, but I often forget to account for more than one character having max frequency .. need to be more careful
+Solved on LeetCode - Yes
+
+
+*/
+
+
+
+
+class Solution {
+    public:
+        int leastInterval(vector<char>& tasks, int n) {
+            int freq_table[26] = {0};
+
+            int totalTasks = tasks.size();
+
+            int maxFreq = INT_MIN;
+
+            for (char c: tasks) {
+                freq_table[c - 'A'] += 1;
+                maxFreq = max(maxFreq, freq_table[c - 'A']);
+            }
+
+            int maxFreqCount = 0;
+            for (int i = 0; i < 26; i++) {
+                if (freq_table[i] == maxFreq) {
+                    maxFreqCount++;
+                }
+            }
+
+            int idle_slots = (maxFreq - 1) * (n - (maxFreqCount - 1));
+
+            idle_slots = idle_slots - (totalTasks - (maxFreq*maxFreqCount));
+            if (idle_slots < 0) idle_slots = 0;
+
+            return totalTasks + idle_slots; 
+
+        }
+    };

02-queue-reconstruction-by-height.cpp

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+/*
+TC - O(N^2) - performed sorting operation O(NlogN), but insertion can be quadratic worst case.
+SC - O(n) - used extra space to hold sorted person objects
+Issues Faced - need to be clear about why we sort like this - it is because without the greedy approach, we would overwrite entries for previously placed people
+Solved on LeetCode - Yes
+
+
+*/
+
+
+class Solution {
+    public:
+        vector<vector<int>> reconstructQueue(vector<vector<int>>& source) {
+
+            std::vector<std::pair<int, int>> target;
+            target.reserve(source.size()); 
+
+            for (const auto &vec : source) {
+                if (vec.size() == 2) {
+                    target.emplace_back(vec[0], vec[1]);
+                }
+            }
+
+            // 2. Sort: Descending by 2nd element, Ascending by 1st element on ties
+            std::sort(target.begin(), target.end(), [](const auto& a, const auto& b) {
+                if (a.first != b.first) {
+                    return a.first > b.first; // Primary sort: Descending heights
+                }
+                return a.second < b.second; // Tie-breaker: Ascending number of people ahead of you (to really put the tallest first)
+            });
+
+            vector<vector<int>> result;
+            for (const auto& person : target) {
+                int height = person.first;
+                int count = person.second;
+
+                // Greedily insert directly at the position dictated by 'count'
+                result.insert(result.begin() + count, {height, count});
+            }
+            return result;
+
+
+        }
+    };

03-partition-labels.cpp

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+/*
+TC - O(n) - two linear passes only
+SC - O(n) - used extra space to hold rightmost positions
+Issues Faced - fairly intuitive, but need to remember two-pass is just one of several greedy string approaches
+Solved on LeetCode - Yes
+
+
+*/
+
+
+
+class Solution {
+    public:
+
+        vector<int> partitionLabels(string s) {
+
+
+
+            int posFromRight[26];
+            std::fill(std::begin(posFromRight), std::end(posFromRight), -1);
+
+
+            for (int i = s.size() - 1 ; i >= 0 ; i--) {
+                int currVal = s[i] - 'a';
+                if (posFromRight[currVal] == -1) {
+                    posFromRight[currVal] = i;
+                    // cout<<"rightmost pos of "<<s[i]<<" is "<<posFromRight[currVal]<<"\n";
+                }
+            }
+
+            int horizon = INT_MIN;
+            int lastEnd = 0;
+            vector<int> res;
+
+            for (int i = 0 ; i < s.size() ; i++) {
+                int currVal = s[i] - 'a';
+                // cout<<"s[i] = "<<s[i]<<"\n";
+                int rightPos = posFromRight[currVal];
+
+                horizon = max(horizon, rightPos);
+
+                if (i == horizon) {
+                    res.push_back(i - lastEnd + 1);
+                    lastEnd = i + 1;
+                }
+
+                // cout<<"horizon is now = "<<horizon<<" for s[i] = "<<s[i]<<"\n";
+            }
+
+            return res;
+
+        }
+    };