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Commitb5c9395

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add 1966
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‎README.md

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@@ -44,6 +44,7 @@ _If you like this project, please leave me a star._ ★
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|1971|[Find if Path Exists in Graph](https://leetcode.com/problems/find-if-path-exists-in-graph/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1971.java)||Easy|BFS, DFS, Graph|
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|1968|[Array With Elements Not Equal to Average of Neighbors](https://leetcode.com/problems/array-with-elements-not-equal-to-average-of-neighbors/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1968.java)||Medium||
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|1967|[Number of Strings That Appear as Substrings in Word](https://leetcode.com/problems/number-of-strings-that-appear-as-substrings-in-word/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1967.java)||Easy||
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|1966|[Binary Searchable Numbers in an Unsorted Array](https://leetcode.com/problems/binary-searchable-numbers-in-an-unsorted-array/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1966.java)||Medium|Array, Binary Search|
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|1961|[Check If String Is a Prefix of Array](https://leetcode.com/problems/check-if-string-is-a-prefix-of-array/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1961.java)||Easy||
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|1957|[Delete Characters to Make Fancy String](https://leetcode.com/problems/delete-characters-to-make-fancy-string/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1957.java)||Easy|String|
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|1952|[Three Divisors](https://leetcode.com/problems/three-divisors/)|[Solution](../master/src/main/java/com/fishercoder/solutions/_1952.java)||Easy||
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packagecom.fishercoder.solutions;
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importjava.util.Deque;
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importjava.util.LinkedList;
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publicclass_1966 {
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publicstaticclassSolution1 {
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/**
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* Brute force: this ends in TLE on LeetCode.
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* The idea is: for every single number in the array, check if there's any number on its right side that's smaller than it
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* and if there's any number on its left side that's bigger than it.
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* If so, based on binary search, if that number gets picked as a pivot for the next search, then this number might not be found.
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*/
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publicintbinarySearchableNumbers(int[]nums) {
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intans =0;
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for (inti =0;i <nums.length;i++) {
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intj =i +1;
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for (;j <nums.length;j++) {
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if (nums[j] <nums[i]) {
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break;
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}
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}
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if (j ==nums.length) {
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intk =i -1;
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for (;k >=0;k--) {
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if (nums[i] <nums[k]) {
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break;
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}
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}
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if (k <=0) {
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ans++;
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}
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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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publicstaticclassSolution2 {
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/**
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* My completely original solution.
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*/
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publicintbinarySearchableNumbers(int[]nums) {
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intans =0;
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int[]maxLeft =newint[nums.length];
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for (inti =0;i <nums.length;i++) {
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maxLeft[i] =i ==0 ?nums[i] :Math.max(nums[i],maxLeft[i -1]);
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}
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int[]minRight =newint[nums.length];
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for (inti =nums.length -1;i >=0;i--) {
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minRight[i] =i +1 ==nums.length ?nums[i] :Math.min(minRight[i +1],nums[i]);
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}
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for (inti =0;i <nums.length;i++) {
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if (nums[i] >=maxLeft[i] &&nums[i] <=minRight[i]) {
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ans++;
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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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publicstaticclassSolution3 {
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/**
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* Using monotonic stack:
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* 1. we only add the ones that are greater than those already on the stack onto the stack.
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* 2. if the existing ones on the stack are greater than the current one,
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* pop them off because they won't be found based on binary search, as a smaller element is on their right side.
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*/
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publicintbinarySearchableNumbers(int[]nums) {
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Deque<Integer>stack =newLinkedList<>();
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intmaxLeft =Integer.MIN_VALUE;
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for (intnum :nums) {
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while (!stack.isEmpty() &&stack.peekLast() >num) {
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stack.pollLast();
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}
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if (num >=maxLeft) {
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stack.addLast(num);
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}
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maxLeft =Math.max(maxLeft,num);
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}
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returnstack.size();
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}
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}
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}
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packagecom.fishercoder;
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importcom.fishercoder.solutions._1966;
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importorg.junit.BeforeClass;
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importorg.junit.Test;
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importstaticorg.junit.Assert.assertEquals;
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publicclass_1966Test {
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privatestatic_1966.Solution1solution1;
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privatestatic_1966.Solution2solution2;
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privatestatic_1966.Solution3solution3;
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privatestaticint[]nums;
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privatestaticintexpected;
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@BeforeClass
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publicstaticvoidsetup() {
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solution1 =new_1966.Solution1();
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solution2 =new_1966.Solution2();
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solution3 =new_1966.Solution3();
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}
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@Test
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publicvoidtest1() {
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nums =newint[]{7};
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expected =1;
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assertEquals(expected,solution1.binarySearchableNumbers(nums));
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assertEquals(expected,solution2.binarySearchableNumbers(nums));
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assertEquals(expected,solution3.binarySearchableNumbers(nums));
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}
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@Test
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publicvoidtest2() {
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nums =newint[]{-1,5,2};
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expected =1;
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assertEquals(expected,solution1.binarySearchableNumbers(nums));
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assertEquals(expected,solution2.binarySearchableNumbers(nums));
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assertEquals(expected,solution3.binarySearchableNumbers(nums));
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}
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@Test
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publicvoidtest3() {
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/**This is to answer the follow-up question, what if duplicates are allowed in the input*/
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nums =newint[]{-1, -1,5,2};
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expected =2;
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assertEquals(expected,solution1.binarySearchableNumbers(nums));
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assertEquals(expected,solution2.binarySearchableNumbers(nums));
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assertEquals(expected,solution3.binarySearchableNumbers(nums));
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}
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@Test
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publicvoidtest4() {
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/**This is to answer the follow-up question, what if duplicates are allowed in the input*/
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nums =newint[]{-1, -1,5,2,2,5};
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expected =3;
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assertEquals(expected,solution1.binarySearchableNumbers(nums));
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assertEquals(expected,solution2.binarySearchableNumbers(nums));
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assertEquals(expected,solution3.binarySearchableNumbers(nums));
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}
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}

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