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Commitd0d71a3

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Added tasks 2785-2789
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‎src/main/java/g2601_2700/s2699_modify_graph_edge_weights/Solution.java‎

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importjava.util.List;
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importjava.util.PriorityQueue;
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@SuppressWarnings("java:S135")
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@SuppressWarnings({"unchecked","java:S135"})
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publicclassSolution {
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publicint[][]modifiedGraphEdges(
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intn,int[][]edges,intsource,intdestination,inttarget) {
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packageg2701_2800.s2785_sort_vowels_in_a_string;
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// #Medium #String #Sorting #2023_09_15_Time_5_ms_(100.00%)_Space_44.6_MB_(96.53%)
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publicclassSolution {
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publicStringsortVowels(Strings) {
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int[]vowelCount =newint[11];
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int[]countIndexMap =newint[128];
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char[]result =s.toCharArray();
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char[]charMap ="AEIOUaeiou".toCharArray();
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for (inti =0;i <charMap.length;i++) {
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countIndexMap[charMap[i]] =i +1;
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}
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for (charc :result) {
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vowelCount[countIndexMap[c]]++;
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}
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intj =1;
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inti =0;
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while (j <vowelCount.length) {
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if (vowelCount[j] >0) {
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while (i <result.length) {
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if (countIndexMap[result[i]] ==0) {
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i++;
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}else {
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vowelCount[j]--;
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result[i++] =charMap[j -1];
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break;
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}
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}
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}else {
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j++;
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}
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}
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returnnewString(result);
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}
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}
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2785\. Sort Vowels in a String
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Medium
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Given a**0-indexed** string`s`,**permute**`s` to get a new string`t` such that:
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* All consonants remain in their original places. More formally, if there is an index`i` with`0 <= i < s.length` such that`s[i]` is a consonant, then`t[i] = s[i]`.
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* The vowels must be sorted in the**nondecreasing** order of their**ASCII** values. More formally, for pairs of indices`i`,`j` with`0 <= i < j < s.length` such that`s[i]` and`s[j]` are vowels, then`t[i]` must not have a higher ASCII value than`t[j]`.
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Return_the resulting string_.
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The vowels are`'a'`,`'e'`,`'i'`,`'o'`, and`'u'`, and they can appear in lowercase or uppercase. Consonants comprise all letters that are not vowels.
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**Example 1:**
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**Input:** s = "lEetcOde"
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**Output:** "lEOtcede"
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**Explanation:** 'E', 'O', and 'e' are the vowels in s; 'l', 't', 'c', and 'd' are all consonants. The vowels are sorted according to their ASCII values, and the consonants remain in the same places.
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**Example 2:**
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**Input:** s = "lYmpH"
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**Output:** "lYmpH"
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**Explanation:** There are no vowels in s (all characters in s are consonants), so we return "lYmpH".
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**Constraints:**
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* <code>1 <= s.length <= 10<sup>5</sup></code>
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*`s` consists only of letters of the English alphabet in**uppercase and lowercase**.
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packageg2701_2800.s2786_visit_array_positions_to_maximize_score;
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// #Medium #Array #Dynamic_Programming #2023_09_15_Time_5_ms_(100.00%)_Space_54.5_MB_(99.95%)
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publicclassSolution {
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publiclongmaxScore(int[]nums,intx) {
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long[]dp = {-x, -x};
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dp[nums[0] &1] =nums[0];
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for (inti =1;i <nums.length;i++) {
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longtoggle =dp[nums[i] &1 ^1] -x;
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longnottoggle =dp[nums[i] &1];
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dp[nums[i] &1] =Math.max(toggle,nottoggle) +nums[i];
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}
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returnMath.max(dp[0],dp[1]);
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}
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}
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2786\. Visit Array Positions to Maximize Score
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Medium
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You are given a**0-indexed** integer array`nums` and a positive integer`x`.
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You are**initially** at position`0` in the array and you can visit other positions according to the following rules:
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* If you are currently in position`i`, then you can move to**any** position`j` such that`i < j`.
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* For each position`i` that you visit, you get a score of`nums[i]`.
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* If you move from a position`i` to a position`j` and the**parities** of`nums[i]` and`nums[j]` differ, then you lose a score of`x`.
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Return_the**maximum** total score you can get_.
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**Note** that initially you have`nums[0]` points.
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**Example 1:**
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**Input:** nums =[2,3,6,1,9,2], x = 5
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**Output:** 13
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**Explanation:**
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We can visit the following positions in the array: 0 -> 2 -> 3 -> 4.
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The corresponding values are 2, 6, 1 and 9. Since the integers 6 and 1 have different parities, the move 2 -> 3 will make you lose a score of x = 5.
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The total score will be: 2 + 6 + 1 + 9 - 5 = 13.
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**Example 2:**
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**Input:** nums =[2,4,6,8], x = 3
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**Output:** 20
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**Explanation:**
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All the integers in the array have the same parities, so we can visit all of them without losing any score.
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The total score is: 2 + 4 + 6 + 8 = 20.
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**Constraints:**
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* <code>2 <= nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i], x <= 10<sup>6</sup></code>
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packageg2701_2800.s2787_ways_to_express_an_integer_as_sum_of_powers;
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// #Medium #Dynamic_Programming #2023_09_15_Time_12_ms_(98.96%)_Space_42.4_MB_(86.68%)
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publicclassSolution {
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publicintnumberOfWays(intn,intx) {
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int[]dp =newint[301];
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intmod =1000000007;
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intv;
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dp[0] =1;
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inta =1;
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while (Math.pow(a,x) <=n) {
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v = (int)Math.pow(a,x);
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for (inti =n;i >=v;i--) {
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dp[i] = (dp[i] +dp[i -v]) %mod;
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}
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a++;
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}
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returndp[n];
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}
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}
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2787\. Ways to Express an Integer as Sum of Powers
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Medium
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Given two**positive** integers`n` and`x`.
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Return_the number of ways_`n`_can be expressed as the sum of the_ <code>x<sup>th</sup></code>_power of**unique** positive integers, in other words, the number of sets of unique integers_ <code>[n<sub>1</sub>, n<sub>2</sub>, ..., n<sub>k</sub>]</code>_where_ <code>n = n<sub>1</sub><sup>x</sup> + n<sub>2</sub><sup>x</sup> + ... + n<sub>k</sub><sup>x</sup></code>_._
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Since the result can be very large, return it modulo <code>10<sup>9</sup> + 7</code>.
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For example, if`n = 160` and`x = 3`, one way to express`n` is <code>n = 2<sup>3</sup> + 3<sup>3</sup> + 5<sup>3</sup></code>.
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**Example 1:**
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**Input:** n = 10, x = 2
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**Output:** 1
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**Explanation:**
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We can express n as the following: n = 3<sup>2</sup> + 1<sup>2</sup> = 10.
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It can be shown that it is the only way to express 10 as the sum of the 2<sup>nd</sup> power of unique integers.
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**Example 2:**
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**Input:** n = 4, x = 1
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**Output:** 2
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**Explanation:**
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We can express n in the following ways:
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- n = 4<sup>1</sup> = 4.
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- n = 3<sup>1</sup> + 1<sup>1</sup> = 4.
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**Constraints:**
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*`1 <= n <= 300`
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*`1 <= x <= 5`
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packageg2701_2800.s2788_split_strings_by_separator;
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// #Easy #Array #String #2023_09_15_Time_3_ms_(99.98%)_Space_44.8_MB_(68.11%)
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importjava.util.ArrayList;
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importjava.util.List;
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publicclassSolution {
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publicList<String>splitWordsBySeparator(List<String>words,charseparator) {
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List<String>list =newArrayList<>();
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for (Stringstr :words) {
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intsi =0;
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for (inti =0;i <str.length();i++) {
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if (str.charAt(i) ==separator) {
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if (i >si) {
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list.add(str.substring(si,i));
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}
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si =i +1;
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}
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}
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if (si !=str.length()) {
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list.add(str.substring(si,str.length()));
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}
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}
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returnlist;
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}
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}
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2788\. Split Strings by Separator
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Easy
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Given an array of strings`words` and a character`separator`,**split** each string in`words` by`separator`.
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Return_an array of strings containing the new strings formed after the splits,**excluding empty strings**._
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**Notes**
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*`separator` is used to determine where the split should occur, but it is not included as part of the resulting strings.
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* A split may result in more than two strings.
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* The resulting strings must maintain the same order as they were initially given.
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**Example 1:**
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**Input:** words =["one.two.three","four.five","six"], separator = "."
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**Output:**["one","two","three","four","five","six"]
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**Explanation:**
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In this example we split as follows: "one.two.three" splits into
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"one", "two", "three" "four.five" splits into
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"four", "five" "six" splits into "six"
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Hence, the resulting array is["one","two","three","four","five","six"].
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**Example 2:**
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**Input:** words =["$easy$","$problem$"], separator = "$"
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**Output:**["easy","problem"]
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**Explanation:**
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In this example we split as follows:
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"$easy$" splits into "easy" (excluding empty strings)
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"$problem$" splits into "problem" (excluding empty strings)
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Hence, the resulting array is["easy","problem"].
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**Example 3:**
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**Input:** words =["|||"], separator = "|"
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**Output:**[]
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**Explanation:**
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In this example the resulting split of "|||" will contain only empty strings, so we return an empty array[].
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**Constraints:**
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*`1 <= words.length <= 100`
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*`1 <= words[i].length <= 20`
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* characters in`words[i]` are either lowercase English letters or characters from the string`".,|$#@"` (excluding the quotes)
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*`separator` is a character from the string`".,|$#@"` (excluding the quotes)
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packageg2701_2800.s2789_largest_element_in_an_array_after_merge_operations;
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// #Medium #Array #Greedy #Prefix_Sum #2023_09_15_Time_1_ms_(100.00%)_Space_58_MB_(58.92%)
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publicclassSolution {
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publiclongmaxArrayValue(int[]nums) {
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longans =nums[nums.length -1];
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for (inti =nums.length -1;i >0; --i) {
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if (ans >=nums[i -1]) {
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ans +=nums[i -1];
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}else {
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ans =nums[i -1];
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}
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}
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returnans;
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}
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}

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