package com.freetymekiyan.algorithms.level.easy;

/**
 * Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
 * <p>
 * Note:
 * You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from
 * nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.
 * <p>
 * Company Tags: Microsoft, Bloomberg, Facebook
 * Tags: Array, Two Pointers
 * Similar Problems: (E) Merge Two Sorted Lists
 */
public class MergeSortedArray {

    /**
     * Array. Two Pointers.
     * One pointer i at the end of nums1. Another pointer j at the end of nums2.
     * Compare their values and put the larger one at the end of nums1.
     * The index is m+n+1.
     * If m == 0, nums1 is fully merged, merge the rest of nums2.
     */
    private void merge(int nums1[], int m, int nums2[], int n) {
        for (int i = m - 1, j = n - 1, k = m + n - 1; k >= 0 && j >= 0; k--) {
            nums1[k] = (i < 0 || nums1[i] < nums2[j]) ? nums2[j--] : nums1[i--];
        }
    }
}

@@ -47,16 +47,49 @@
 y;%0A%0A/**%0A
+ * 88. Merge Sorted Array%0A * %3Cp%3E%0A
  * Given
@@ -785,11 +785,15 @@
 is m
-+n+
+ + n - 
 1.%0A 
@@ -803,16 +803,20 @@
  * If m 
+- 1 
 == 0, nu
@@ -878,14 +878,13 @@
    p
-rivate
+ublic
  voi
@@ -898,16 +898,16 @@
 (int
+%5B%5D
  nums1
-%5B%5D
 , in
@@ -918,16 +918,16 @@
  int
+%5B%5D
  nums2
-%5B%5D
 , in
@@ -1109,10 +1109,332 @@
 %7D%0A    %7D%0A
-%7D
 %0A
+    /**%0A     * Instead of 3 indices, use m-1 for nums1, n-1 for nums2.%0A     */%0A    public void merge2(int%5B%5D nums1, int m, int%5B%5D nums2, int n) %7B%0A        for (int i = m + n - 1; i %3E= 0 && n - 1 %3E= 0; i--) %7B%0A            nums1%5Bi%5D = (m - 1 %3C 0 %7C%7C nums1%5Bm - 1%5D %3C nums2%5Bn - 1%5D) ? nums2%5Bn-- - 1%5D : nums1%5Bm-- - 1%5D;%0A        %7D%0A    %7D%0A%7D

update 88. Merge Sorted Array fix explanation, add new solution with fewer indices