330 Patching Array

Problem:

Given a sorted positive integer array nums and an integer n, add/patch elements to the array such that any number in range [1, n] inclusive can be formed by the sum of some elements in the array. Return the minimum number of patches required.

Example 1: nums = [1, 3], n = 6 Return 1.

Combinations of nums are [1], [3], [1,3], which form possible sums of: 1, 3, 4. Now if we add/patch 2 to nums, the combinations are: [1], [2], [3], [1,3], [2,3], [1,2,3]. Possible sums are 1, 2, 3, 4, 5, 6, which now covers the range [1, 6]. So we only need 1 patch.

Example 2: nums = [1, 5, 10], n = 20 Return 2. The two patches can be [2, 4].

Example 3: nums = [1, 2, 2], n = 5 Return 0.

Solutions:

Memory exceeded solution

``````public class Solution {
public int minPatches(int[] nums, int n) {
boolean[] has = new boolean[n];
int[] toFill = new int[1];
toFill[0] = n;
for (int i = 0; i < nums.length; i ++) {
if (nums[i] <= n) {
}
}
int start = 0;
int count = 0;
while (toFill[0] != 0) {
while (has[start] == true) {
start ++;
}
addElement(start + 1, has, nums, toFill, n);
count ++;
}
return count;
}
private void addElement(int add, boolean[] has, int[] nums, int[] toFill, int n) {
for (int j = 0; j < n; j ++) {
if (has[j] == true && j + add < n && has[j + add] == false) {
}
}
toFill[0] --;
}
if (has[add - 1] == false) {
toFill[0] --;
}
}
}
``````

Greedy working solution

``````public class Solution {
public int minPatches(int[] nums, int n) {
long miss = 1;
int count = 0;
int i = 0;
while (miss <= n) {
if (i < nums.length && nums[i] <= miss) {
miss += nums[i];
i ++;
}
else {
miss += miss;
count ++;
}
}
return count;
}
}
``````