Added tasks 3142-3149

Author: javadev

src/main/java/g3101_3200/s3142_check_if_grid_satisfies_conditions/Solution.java

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+package g3101_3200.s3142_check_if_grid_satisfies_conditions;
+
+// #Easy #Array #Matrix #2024_05_15_Time_1_ms_(95.76%)_Space_44.4_MB_(59.70%)
+
+public class Solution {
+    public boolean satisfiesConditions(int[][] grid) {
+        int m = grid.length;
+        int n = grid[0].length;
+        for (int i = 0; i < m - 1; i++) {
+            if (n > 1) {
+                for (int j = 0; j < n - 1; j++) {
+                    if ((grid[i][j] != grid[i + 1][j]) || (grid[i][j] == grid[i][j + 1])) {
+                        return false;
+                    }
+                }
+            } else {
+                if (grid[i][0] != grid[i + 1][0]) {
+                    return false;
+                }
+            }
+        }
+        for (int j = 0; j < n - 1; j++) {
+            if (grid[m - 1][j] == grid[m - 1][j + 1]) {
+                return false;
+            }
+        }
+        return true;
+    }
+}

src/main/java/g3101_3200/s3142_check_if_grid_satisfies_conditions/readme.md

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+3142\. Check if Grid Satisfies Conditions
+
+Easy
+
+You are given a 2D matrix `grid` of size `m x n`. You need to check if each cell `grid[i][j]` is:
+
+*   Equal to the cell below it, i.e. `grid[i][j] == grid[i + 1][j]` (if it exists).
+*   Different from the cell to its right, i.e. `grid[i][j] != grid[i][j + 1]` (if it exists).
+
+Return `true` if **all** the cells satisfy these conditions, otherwise, return `false`.
+
+**Example 1:**
+
+**Input:** grid = [[1,0,2],[1,0,2]]
+
+**Output:** true
+
+**Explanation:**
+
+**![](https://assets.leetcode.com/uploads/2024/04/15/examplechanged.png)**
+
+All the cells in the grid satisfy the conditions.
+
+**Example 2:**
+
+**Input:** grid = [[1,1,1],[0,0,0]]
+
+**Output:** false
+
+**Explanation:**
+
+**![](https://assets.leetcode.com/uploads/2024/03/27/example21.png)**
+
+All cells in the first row are equal.
+
+**Example 3:**
+
+**Input:** grid = [[1],[2],[3]]
+
+**Output:** false
+
+**Explanation:**
+
+![](https://assets.leetcode.com/uploads/2024/03/31/changed.png)
+
+Cells in the first column have different values.
+
+**Constraints:**
+
+*   `1 <= n, m <= 10`
+*   `0 <= grid[i][j] <= 9`

src/main/java/g3101_3200/s3143_maximum_points_inside_the_square/Solution.java

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+package g3101_3200.s3143_maximum_points_inside_the_square;
+
+// #Medium #Array #String #Hash_Table #Sorting #Binary_Search
+// #2024_05_15_Time_2_ms_(100.00%)_Space_100.1_MB_(61.27%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public int maxPointsInsideSquare(int[][] points, String s) {
+        int[] tags = new int[26];
+        Arrays.fill(tags, Integer.MAX_VALUE);
+        int secondMin = Integer.MAX_VALUE;
+        for (int i = 0; i < s.length(); i++) {
+            int dist = Math.max(Math.abs(points[i][0]), Math.abs(points[i][1]));
+            char c = s.charAt(i);
+            if (tags[c - 'a'] == Integer.MAX_VALUE) {
+                tags[c - 'a'] = dist;
+            } else if (dist < tags[c - 'a']) {
+                secondMin = Math.min(secondMin, tags[c - 'a']);
+                tags[c - 'a'] = dist;
+            } else {
+                secondMin = Math.min(secondMin, dist);
+            }
+        }
+        int count = 0;
+        for (int dist : tags) {
+            if (dist < secondMin) {
+                count++;
+            }
+        }
+        return count;
+    }
+}

src/main/java/g3101_3200/s3143_maximum_points_inside_the_square/readme.md

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+3143\. Maximum Points Inside the Square
+
+Medium
+
+You are given a 2D array `points` and a string `s` where, `points[i]` represents the coordinates of point `i`, and `s[i]` represents the **tag** of point `i`.
+
+A **valid** square is a square centered at the origin `(0, 0)`, has edges parallel to the axes, and **does not** contain two points with the same tag.
+
+Return the **maximum** number of points contained in a **valid** square.
+
+Note:
+
+*   A point is considered to be inside the square if it lies on or within the square's boundaries.
+*   The side length of the square can be zero.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2024/03/29/3708-tc1.png)
+
+**Input:** points = [[2,2],[-1,-2],[-4,4],[-3,1],[3,-3]], s = "abdca"
+
+**Output:** 2
+
+**Explanation:**
+
+The square of side length 4 covers two points `points[0]` and `points[1]`.
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2024/03/29/3708-tc2.png)
+
+**Input:** points = [[1,1],[-2,-2],[-2,2]], s = "abb"
+
+**Output:** 1
+
+**Explanation:**
+
+The square of side length 2 covers one point, which is `points[0]`.
+
+**Example 3:**
+
+**Input:** points = [[1,1],[-1,-1],[2,-2]], s = "ccd"
+
+**Output:** 0
+
+**Explanation:**
+
+It's impossible to make any valid squares centered at the origin such that it covers only one point among `points[0]` and `points[1]`.
+
+**Constraints:**
+
+*   <code>1 <= s.length, points.length <= 10<sup>5</sup></code>
+*   `points[i].length == 2`
+*   <code>-10<sup>9</sup> <= points[i][0], points[i][1] <= 10<sup>9</sup></code>
+*   `s.length == points.length`
+*   `points` consists of distinct coordinates.
+*   `s` consists only of lowercase English letters.

src/main/java/g3101_3200/s3144_minimum_substring_partition_of_equal_character_frequency/Solution.java

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+package g3101_3200.s3144_minimum_substring_partition_of_equal_character_frequency;
+
+// #Medium #String #Hash_Table #Dynamic_Programming #Counting
+// #2024_05_15_Time_37_ms_(100.00%)_Space_44.9_MB_(72.95%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public int minimumSubstringsInPartition(String s) {
+        char[] cs = s.toCharArray();
+        int n = cs.length;
+        int[] dp = new int[n + 1];
+        Arrays.fill(dp, n);
+        dp[0] = 0;
+        for (int i = 1; i <= n; ++i) {
+            int[] count = new int[26];
+            int distinct = 0;
+            int maxCount = 0;
+            for (int j = i - 1; j >= 0; --j) {
+                int index = cs[j] - 'a';
+                if (++count[index] == 1) {
+                    distinct++;
+                }
+                if (count[index] > maxCount) {
+                    maxCount = count[index];
+                }
+                if (maxCount * distinct == i - j) {
+                    dp[i] = Math.min(dp[i], dp[j] + 1);
+                }
+            }
+        }
+        return dp[n];
+    }
+}

src/main/java/g3101_3200/s3144_minimum_substring_partition_of_equal_character_frequency/readme.md

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+3144\. Minimum Substring Partition of Equal Character Frequency
+
+Medium
+
+Given a string `s`, you need to partition it into one or more **balanced** substrings. For example, if `s == "ababcc"` then `("abab", "c", "c")`, `("ab", "abc", "c")`, and `("ababcc")` are all valid partitions, but <code>("a", **"bab"**, "cc")</code>, <code>(**"aba"**, "bc", "c")</code>, and <code>("ab", **"abcc"**)</code> are not. The unbalanced substrings are bolded.
+
+Return the **minimum** number of substrings that you can partition `s` into.
+
+**Note:** A **balanced** string is a string where each character in the string occurs the same number of times.
+
+**Example 1:**
+
+**Input:** s = "fabccddg"
+
+**Output:** 3
+
+**Explanation:**
+
+We can partition the string `s` into 3 substrings in one of the following ways: `("fab, "ccdd", "g")`, or `("fabc", "cd", "dg")`.
+
+**Example 2:**
+
+**Input:** s = "abababaccddb"
+
+**Output:** 2
+
+**Explanation:**
+
+We can partition the string `s` into 2 substrings like so: `("abab", "abaccddb")`.
+
+**Constraints:**
+
+*   `1 <= s.length <= 1000`
+*   `s` consists only of English lowercase letters.

src/main/java/g3101_3200/s3145_find_products_of_elements_of_big_array/Solution.java

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+package g3101_3200.s3145_find_products_of_elements_of_big_array;
+
+// #Hard #Array #Binary_Search #Bit_Manipulation
+// #2024_05_15_Time_3_ms_(98.41%)_Space_44.5_MB_(96.83%)
+
+public class Solution {
+    public int[] findProductsOfElements(long[][] queries) {
+        int[] ans = new int[queries.length];
+        for (int i = 0; i < queries.length; i++) {
+            long[] q = queries[i];
+            long er = sumE(q[1] + 1);
+            long el = sumE(q[0]);
+            ans[i] = pow(2, er - el, q[2]);
+        }
+        return ans;
+    }
+
+    private long sumE(long k) {
+        long res = 0, n = 0, cnt1 = 0, sumI = 0;
+        for (long i = 63 - Long.numberOfLeadingZeros(k + 1); i > 0; i--) {
+            long c = (cnt1 << i) + (i << (i - 1));
+            if (c <= k) {
+                k -= c;
+                res += (sumI << i) + ((i * (i - 1) / 2) << (i - 1));
+                sumI += i;
+                cnt1++;
+                n |= 1L << i;
+            }
+        }
+        if (cnt1 <= k) {
+            k -= cnt1;
+            res += sumI;
+            n++;
+        }
+        while (k-- > 0) {
+            res += Long.numberOfTrailingZeros(n);
+            n &= n - 1;
+        }
+        return res;
+    }
+
+    private int pow(long x, long n, long mod) {
+        long res = 1 % mod;
+        for (; n > 0; n /= 2) {
+            if (n % 2 == 1) {
+                res = (res * x) % mod;
+            }
+            x = (x * x) % mod;
+        }
+        return (int) res;
+    }
+}

src/main/java/g3101_3200/s3145_find_products_of_elements_of_big_array/readme.md

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+3145\. Find Products of Elements of Big Array
+
+Hard
+
+A **powerful array** for an integer `x` is the shortest sorted array of powers of two that sum up to `x`. For example, the powerful array for 11 is `[1, 2, 8]`.
+
+The array `big_nums` is created by concatenating the **powerful** arrays for every positive integer `i` in ascending order: 1, 2, 3, and so forth. Thus, `big_nums` starts as <code>[<ins>1</ins>, <ins>2</ins>, <ins>1, 2</ins>, <ins>4</ins>, <ins>1, 4</ins>, <ins>2, 4</ins>, <ins>1, 2, 4</ins>, <ins>8</ins>, ...]</code>.
+
+You are given a 2D integer matrix `queries`, where for <code>queries[i] = [from<sub>i</sub>, to<sub>i</sub>, mod<sub>i</sub>]</code> you should calculate <code>(big_nums[from<sub>i</sub>] * big_nums[from<sub>i</sub> + 1] * ... * big_nums[to<sub>i</sub>]) % mod<sub>i</sub></code>.
+
+Return an integer array `answer` such that `answer[i]` is the answer to the <code>i<sup>th</sup></code> query.
+
+**Example 1:**
+
+**Input:** queries = [[1,3,7]]
+
+**Output:** [4]
+
+**Explanation:**
+
+There is one query.
+
+`big_nums[1..3] = [2,1,2]`. The product of them is 4. The remainder of 4 under 7 is 4.
+
+**Example 2:**
+
+**Input:** queries = [[2,5,3],[7,7,4]]
+
+**Output:** [2,2]
+
+**Explanation:**
+
+There are two queries.
+
+First query: `big_nums[2..5] = [1,2,4,1]`. The product of them is 8. The remainder of 8 under 3 is 2.
+
+Second query: `big_nums[7] = 2`. The remainder of 2 under 4 is 2.
+
+**Constraints:**
+
+*   `1 <= queries.length <= 500`
+*   `queries[i].length == 3`
+*   <code>0 <= queries[i][0] <= queries[i][1] <= 10<sup>15</sup></code>
+*   <code>1 <= queries[i][2] <= 10<sup>5</sup></code>

src/main/java/g3101_3200/s3146_permutation_difference_between_two_strings/Solution.java

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+package g3101_3200.s3146_permutation_difference_between_two_strings;
+
+// #Easy #String #Hash_Table #2024_05_15_Time_1_ms_(100.00%)_Space_42.4_MB_(84.38%)
+
+import java.util.Arrays;
+
+public class Solution {
+    public int findPermutationDifference(String s, String t) {
+        int[] res = new int[26];
+        Arrays.fill(res, -1);
+        int sum = 0;
+        for (int i = 0; i < s.length(); ++i) {
+            res[s.charAt(i) - 'a'] = i;
+        }
+        for (int i = 0; i < t.length(); ++i) {
+            sum += Math.abs(res[t.charAt(i) - 'a'] - i);
+        }
+        return sum;
+    }
+}