Added tasks 3184-3187

Author: javadev

src/main/java/g3101_3200/s3184_count_pairs_that_form_a_complete_day_i/Solution.java

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+package g3101_3200.s3184_count_pairs_that_form_a_complete_day_i;
+
+// #Easy #Array #Hash_Table #Counting #2024_06_21_Time_1_ms_(98.20%)_Space_42_MB_(83.72%)
+
+public class Solution {
+    public int countCompleteDayPairs(int[] hours) {
+        int[] modular = new int[26];
+        int ans = 0;
+        for (int hour : hours) {
+            int mod = hour % 24;
+            ans += modular[24 - mod];
+            if (mod == 0) {
+                modular[24]++;
+            } else {
+                modular[mod]++;
+            }
+        }
+        return ans;
+    }
+}

src/main/java/g3101_3200/s3184_count_pairs_that_form_a_complete_day_i/readme.md

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+3184\. Count Pairs That Form a Complete Day I
+
+Easy
+
+Given an integer array `hours` representing times in **hours**, return an integer denoting the number of pairs `i`, `j` where `i < j` and `hours[i] + hours[j]` forms a **complete day**.
+
+A **complete day** is defined as a time duration that is an **exact** **multiple** of 24 hours.
+
+For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.
+
+**Example 1:**
+
+**Input:** hours = [12,12,30,24,24]
+
+**Output:** 2
+
+**Explanation:**
+
+The pairs of indices that form a complete day are `(0, 1)` and `(3, 4)`.
+
+**Example 2:**
+
+**Input:** hours = [72,48,24,3]
+
+**Output:** 3
+
+**Explanation:**
+
+The pairs of indices that form a complete day are `(0, 1)`, `(0, 2)`, and `(1, 2)`.
+
+**Constraints:**
+
+*   `1 <= hours.length <= 100`
+*   <code>1 <= hours[i] <= 10<sup>9</sup></code>

src/main/java/g3101_3200/s3185_count_pairs_that_form_a_complete_day_ii/Solution.java

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+package g3101_3200.s3185_count_pairs_that_form_a_complete_day_ii;
+
+// #Medium #Array #Hash_Table #Counting #2024_06_21_Time_3_ms_(97.60%)_Space_97.1_MB_(14.69%)
+
+public class Solution {
+    public long countCompleteDayPairs(int[] hours) {
+        long[] hour = new long[24];
+        for (int j : hours) {
+            hour[j % 24]++;
+        }
+        long counter = hour[0] * (hour[0] - 1) / 2;
+        counter += hour[12] * (hour[12] - 1) / 2;
+        for (int i = 1; i < 12; i++) {
+            counter += hour[i] * hour[24 - i];
+        }
+        return counter;
+    }
+}

src/main/java/g3101_3200/s3185_count_pairs_that_form_a_complete_day_ii/readme.md

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+3185\. Count Pairs That Form a Complete Day II
+
+Medium
+
+Given an integer array `hours` representing times in **hours**, return an integer denoting the number of pairs `i`, `j` where `i < j` and `hours[i] + hours[j]` forms a **complete day**.
+
+A **complete day** is defined as a time duration that is an **exact** **multiple** of 24 hours.
+
+For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.
+
+**Example 1:**
+
+**Input:** hours = [12,12,30,24,24]
+
+**Output:** 2
+
+**Explanation:** The pairs of indices that form a complete day are `(0, 1)` and `(3, 4)`.
+
+**Example 2:**
+
+**Input:** hours = [72,48,24,3]
+
+**Output:** 3
+
+**Explanation:** The pairs of indices that form a complete day are `(0, 1)`, `(0, 2)`, and `(1, 2)`.
+
+**Constraints:**
+
+*   <code>1 <= hours.length <= 5 * 10<sup>5</sup></code>
+*   <code>1 <= hours[i] <= 10<sup>9</sup></code>

src/main/java/g3101_3200/s3186_maximum_total_damage_with_spell_casting/Solution.java

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+package g3101_3200.s3186_maximum_total_damage_with_spell_casting;
+
+// #Medium #Array #Hash_Table #Dynamic_Programming #Sorting #Binary_Search #Two_Pointers #Counting
+// #2024_06_21_Time_51_ms_(99.29%)_Space_60.8_MB_(78.34%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public long maximumTotalDamage(int[] power) {
+        int maxPower = 0;
+        for (int p : power) {
+            if (p > maxPower) {
+                maxPower = p;
+            }
+        }
+        return (maxPower <= 1_000_000) ? smallPower(power, maxPower) : bigPower(power);
+    }
+
+    private long smallPower(int[] power, int maxPower) {
+        int[] counts = new int[maxPower + 6];
+        for (int p : power) {
+            counts[p]++;
+        }
+        long[] dp = new long[maxPower + 6];
+        dp[1] = counts[1];
+        dp[2] = Math.max(counts[2] * 2L, dp[1]);
+        for (int i = 3; i <= maxPower; i++) {
+            dp[i] = Math.max(counts[i] * i + dp[i - 3], Math.max(dp[i - 1], dp[i - 2]));
+        }
+        return dp[maxPower];
+    }
+
+    private long bigPower(int[] power) {
+        Arrays.sort(power);
+        int n = power.length;
+        long[] prevs = new long[4];
+        int curPower = power[0];
+        int count = 1;
+        long result = 0;
+        for (int i = 1; i <= n; i++) {
+            int p = (i == n) ? 1_000_000_009 : power[i];
+            if (p == curPower) {
+                count++;
+            } else {
+                long curVal =
+                        Math.max((long) curPower * count + prevs[3], Math.max(prevs[1], prevs[2]));
+                int diff = Math.min(p - curPower, prevs.length - 1);
+                long nextCurVal = (diff == 1) ? 0 : Math.max(prevs[3], Math.max(curVal, prevs[2]));
+                // Shift the values in prevs[].
+                int k = prevs.length - 1;
+                if (diff < prevs.length - 1) {
+                    while (k > diff) {
+                        prevs[k] = prevs[k-- - diff];
+                    }
+                    prevs[k--] = curVal;
+                }
+                while (k > 0) {
+                    prevs[k--] = nextCurVal;
+                }
+                curPower = p;
+                count = 1;
+            }
+        }
+        for (long v : prevs) {
+            if (v > result) {
+                result = v;
+            }
+        }
+        return result;
+    }
+}

src/main/java/g3101_3200/s3186_maximum_total_damage_with_spell_casting/readme.md

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+3186\. Maximum Total Damage With Spell Casting
+
+Medium
+
+A magician has various spells.
+
+You are given an array `power`, where each element represents the damage of a spell. Multiple spells can have the same damage value.
+
+It is a known fact that if a magician decides to cast a spell with a damage of `power[i]`, they **cannot** cast any spell with a damage of `power[i] - 2`, `power[i] - 1`, `power[i] + 1`, or `power[i] + 2`.
+
+Each spell can be cast **only once**.
+
+Return the **maximum** possible _total damage_ that a magician can cast.
+
+**Example 1:**
+
+**Input:** power = [1,1,3,4]
+
+**Output:** 6
+
+**Explanation:**
+
+The maximum possible damage of 6 is produced by casting spells 0, 1, 3 with damage 1, 1, 4.
+
+**Example 2:**
+
+**Input:** power = [7,1,6,6]
+
+**Output:** 13
+
+**Explanation:**
+
+The maximum possible damage of 13 is produced by casting spells 1, 2, 3 with damage 1, 6, 6.
+
+**Constraints:**
+
+*   <code>1 <= power.length <= 10<sup>5</sup></code>
+*   <code>1 <= power[i] <= 10<sup>9</sup></code>

src/main/java/g3101_3200/s3187_peaks_in_array/Solution.java

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+package g3101_3200.s3187_peaks_in_array;
+
+// #Hard #Array #Segment_Tree #Binary_Indexed_Tree
+// #2024_06_21_Time_18_ms_(100.00%)_Space_137.7_MB_(31.34%)
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class Solution {
+    public List<Integer> countOfPeaks(int[] nums, int[][] queries) {
+        boolean[] peaks = new boolean[nums.length];
+        int[] binaryIndexedTree = new int[Integer.highestOneBit(peaks.length) * 2 + 1];
+        for (int i = 1; i < peaks.length - 1; ++i) {
+            if (nums[i] > Math.max(nums[i - 1], nums[i + 1])) {
+                peaks[i] = true;
+                update(binaryIndexedTree, i + 1, 1);
+            }
+        }
+
+        List<Integer> result = new ArrayList<>();
+        for (int[] query : queries) {
+            if (query[0] == 1) {
+                int leftIndex = query[1];
+                int rightIndex = query[2];
+                result.add(computeRangeSum(binaryIndexedTree, leftIndex + 2, rightIndex));
+            } else {
+                int index = query[1];
+                int value = query[2];
+                nums[index] = value;
+                for (int i = -1; i <= 1; ++i) {
+                    int affected = index + i;
+                    if (affected >= 1 && affected <= nums.length - 2) {
+                        boolean peak =
+                                nums[affected] > Math.max(nums[affected - 1], nums[affected + 1]);
+                        if (peak != peaks[affected]) {
+                            if (peak) {
+                                update(binaryIndexedTree, affected + 1, 1);
+                            } else {
+                                update(binaryIndexedTree, affected + 1, -1);
+                            }
+                            peaks[affected] = peak;
+                        }
+                    }
+                }
+            }
+        }
+        return result;
+    }
+
+    private int computeRangeSum(int[] binaryIndexedTree, int beginIndex, int endIndex) {
+        return (beginIndex <= endIndex)
+                ? (query(binaryIndexedTree, endIndex) - query(binaryIndexedTree, beginIndex - 1))
+                : 0;
+    }
+
+    private int query(int[] binaryIndexedTree, int index) {
+        int result = 0;
+        while (index != 0) {
+            result += binaryIndexedTree[index];
+            index -= index & -index;
+        }
+
+        return result;
+    }
+
+    private void update(int[] binaryIndexedTree, int index, int delta) {
+        while (index < binaryIndexedTree.length) {
+            binaryIndexedTree[index] += delta;
+            index += index & -index;
+        }
+    }
+}

src/main/java/g3101_3200/s3187_peaks_in_array/readme.md

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+3187\. Peaks in Array
+
+Hard
+
+A **peak** in an array `arr` is an element that is **greater** than its previous and next element in `arr`.
+
+You are given an integer array `nums` and a 2D integer array `queries`.
+
+You have to process queries of two types:
+
+*   <code>queries[i] = [1, l<sub>i</sub>, r<sub>i</sub>]</code>, determine the count of **peak** elements in the subarray <code>nums[l<sub>i</sub>..r<sub>i</sub>]</code>.
+*   <code>queries[i] = [2, index<sub>i</sub>, val<sub>i</sub>]</code>, change <code>nums[index<sub>i</sub>]</code> to <code>val<sub>i</sub></code>.
+
+Return an array `answer` containing the results of the queries of the first type in order.
+
+**Notes:**
+
+*   The **first** and the **last** element of an array or a subarray **cannot** be a peak.
+
+**Example 1:**
+
+**Input:** nums = [3,1,4,2,5], queries = [[2,3,4],[1,0,4]]
+
+**Output:** [0]
+
+**Explanation:**
+
+First query: We change `nums[3]` to 4 and `nums` becomes `[3,1,4,4,5]`.
+
+Second query: The number of peaks in the `[3,1,4,4,5]` is 0.
+
+**Example 2:**
+
+**Input:** nums = [4,1,4,2,1,5], queries = [[2,2,4],[1,0,2],[1,0,4]]
+
+**Output:** [0,1]
+
+**Explanation:**
+
+First query: `nums[2]` should become 4, but it is already set to 4.
+
+Second query: The number of peaks in the `[4,1,4]` is 0.
+
+Third query: The second 4 is a peak in the `[4,1,4,2,1]`.
+
+**Constraints:**
+
+*   <code>3 <= nums.length <= 10<sup>5</sup></code>
+*   <code>1 <= nums[i] <= 10<sup>5</sup></code>
+*   <code>1 <= queries.length <= 10<sup>5</sup></code>
+*   `queries[i][0] == 1` or `queries[i][0] == 2`
+*   For all `i` that:
+    *   `queries[i][0] == 1`: `0 <= queries[i][1] <= queries[i][2] <= nums.length - 1`
+    *   `queries[i][0] == 2`: `0 <= queries[i][1] <= nums.length - 1`, <code>1 <= queries[i][2] <= 10<sup>5</sup></code>

src/test/java/g3101_3200/s3184_count_pairs_that_form_a_complete_day_i/SolutionTest.java

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+package g3101_3200.s3184_count_pairs_that_form_a_complete_day_i;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void countCompleteDayPairs() {
+        assertThat(
+                new Solution().countCompleteDayPairs(new int[] {12, 12, 30, 24, 24}), equalTo(2));
+    }
+
+    @Test
+    void countCompleteDayPairs2() {
+        assertThat(new Solution().countCompleteDayPairs(new int[] {72, 48, 24, 3}), equalTo(3));
+    }
+}

src/test/java/g3101_3200/s3185_count_pairs_that_form_a_complete_day_ii/SolutionTest.java

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+package g3101_3200.s3185_count_pairs_that_form_a_complete_day_ii;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void countCompleteDayPairs() {
+        assertThat(
+                new Solution().countCompleteDayPairs(new int[] {12, 12, 30, 24, 24}), equalTo(2L));
+    }
+
+    @Test
+    void countCompleteDayPairs2() {
+        assertThat(new Solution().countCompleteDayPairs(new int[] {72, 48, 24, 3}), equalTo(3L));
+    }
+}