##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2333.html

# 2333. Minimum Sum of Squared Difference

• Difficulty: Medium.
• Related Topics: Array, Math, Sorting, Heap (Priority Queue).
• Similar Questions: Minimum Absolute Sum Difference, Partition Array Into Two Arrays to Minimize Sum Difference.

## Problem

You are given two positive 0-indexed integer arrays nums1 and nums2, both of length n.

The sum of squared difference of arrays nums1 and nums2 is defined as the sum of (nums1[i] - nums2[i])2 for each 0 <= i < n.

You are also given two positive integers k1 and k2. You can modify any of the elements of nums1 by +1 or -1 at most k1 times. Similarly, you can modify any of the elements of nums2 by +1 or -1 at most k2 times.

Return the minimum **sum of squared difference after modifying array nums1 at most k1 times and modifying array nums2 at most k2 times**.

Note: You are allowed to modify the array elements to become negative integers.

Example 1:

Input: nums1 = [1,2,3,4], nums2 = [2,10,20,19], k1 = 0, k2 = 0
Output: 579
Explanation: The elements in nums1 and nums2 cannot be modified because k1 = 0 and k2 = 0.
The sum of square difference will be: (1 - 2)2 + (2 - 10)2 + (3 - 20)2 + (4 - 19)2 = 579.


Example 2:

Input: nums1 = [1,4,10,12], nums2 = [5,8,6,9], k1 = 1, k2 = 1
Output: 43
Explanation: One way to obtain the minimum sum of square difference is:
- Increase nums1[0] once.
- Increase nums2[2] once.
The minimum of the sum of square difference will be:
(2 - 5)2 + (4 - 8)2 + (10 - 7)2 + (12 - 9)2 = 43.
Note that, there are other ways to obtain the minimum of the sum of square difference, but there is no way to obtain a sum smaller than 43.


Constraints:

• n == nums1.length == nums2.length

• 1 <= n <= 105

• 0 <= nums1[i], nums2[i] <= 105

• 0 <= k1, k2 <= 109

## Solution (Java, C++, Python)

• class Solution {
public long minSumSquareDiff(int[] nums1, int[] nums2, int k1, int k2) {
long minSumSquare = 0;
int[] diffs = new int[100_001];
long totalDiff = 0;
long kSum = (long) k1 + k2;
int currentDiff;
int maxDiff = 0;
for (int i = 0; i < nums1.length; i++) {
// get current diff.
currentDiff = Math.abs(nums1[i] - nums2[i]);
// if current diff > 0, count/store it. If not,then ignore it.
if (currentDiff > 0) {
totalDiff += currentDiff;
diffs[currentDiff]++;
maxDiff = Math.max(maxDiff, currentDiff);
}
}
// if kSum (k1 + k2) < totalDifferences, it means we can make all numbers/differences 0s
if (totalDiff <= kSum) {
return 0;
}
// starting from the back, from the highest difference, lower that group one by one to the
// previous group.
// we need to make all n diffs to n-1, then n-2, as long as kSum allows it.
for (int i = maxDiff; i > 0 && kSum > 0; i--) {
if (diffs[i] > 0) {
// if current group has more differences than the totalK, we can only move k of them
// to the lower level.
if (diffs[i] >= kSum) {
diffs[i] -= kSum;
diffs[i - 1] += kSum;
kSum = 0;
} else {
// else, we can make this whole group one level lower.
diffs[i - 1] += diffs[i];
kSum -= diffs[i];
diffs[i] = 0;
}
}
}
for (int i = 0; i <= maxDiff; i++) {
if (diffs[i] > 0) {
minSumSquare += (long) (Math.pow(i, 2)) * diffs[i];
}
}
return minSumSquare;
}
}

############

class Solution {
public long minSumSquareDiff(int[] nums1, int[] nums2, int k1, int k2) {
int n = nums1.length;
int[] d = new int[n];
long s = 0;
int mx = 0;
int k = k1 + k2;
for (int i = 0; i < n; ++i) {
d[i] = Math.abs(nums1[i] - nums2[i]);
s += d[i];
mx = Math.max(mx, d[i]);
}
if (s <= k) {
return 0;
}
int left = 0, right = mx;
while (left < right) {
int mid = (left + right) >> 1;
long t = 0;
for (int v : d) {
t += Math.max(v - mid, 0);
}
if (t <= k) {
right = mid;
} else {
left = mid + 1;
}
}
for (int i = 0; i < n; ++i) {
k -= Math.max(0, d[i] - left);
d[i] = Math.min(d[i], left);
}
for (int i = 0; i < n && k > 0; ++i) {
if (d[i] == left) {
--k;
--d[i];
}
}
long ans = 0;
for (int v : d) {
ans += (long) v * v;
}
return ans;
}
}

• class Solution:
def minSumSquareDiff(
self, nums1: List[int], nums2: List[int], k1: int, k2: int
) -> int:
d = [abs(a - b) for a, b in zip(nums1, nums2)]
k = k1 + k2
if sum(d) <= k:
return 0
left, right = 0, max(d)
while left < right:
mid = (left + right) >> 1
if sum(max(v - mid, 0) for v in d) <= k:
right = mid
else:
left = mid + 1
for i, v in enumerate(d):
d[i] = min(left, v)
k -= max(0, v - left)
for i, v in enumerate(d):
if k == 0:
break
if v == left:
k -= 1
d[i] -= 1
return sum(v * v for v in d)

############

# 2333. Minimum Sum of Squared Difference
# https://leetcode.com/problems/minimum-sum-of-squared-difference/

class Solution:
def minSumSquareDiff(self, nums1: List[int], nums2: List[int], k1: int, k2: int) -> int:
pq = []

for a, b in zip(nums1, nums2):
heappush(pq, ((-abs(a - b), 1)))

orders = k1 + k2

while pq and orders > 0:
flag = True
top, count = heappop(pq)
top = -top

while pq and top == -pq[0][0]:
count += heappop(pq)[1]
nextTop = 0
if pq:
nextTop = -pq[0][0]

delta = top - nextTop
if delta * count <= orders:
orders -= (delta * count)
else:
flag = False
heappush(pq, (-top, count))

while pq and orders > 0:
x, cnt = heappop(pq)
take = min(orders, cnt)
orders -= take
heappush(pq, (x + 1, take))
if cnt != take:
heappush(pq, (x, cnt - take))

if flag and nextTop:
heappush(pq, (-nextTop, count))

total = 0
for x, count in pq:
total += x * x * count


• using ll = long long;

class Solution {
public:
long long minSumSquareDiff(vector<int>& nums1, vector<int>& nums2, int k1, int k2) {
int n = nums1.size();
vector<int> d(n);
ll s = 0;
int mx = 0;
int k = k1 + k2;
for (int i = 0; i < n; ++i) {
d[i] = abs(nums1[i] - nums2[i]);
s += d[i];
mx = max(mx, d[i]);
}
if (s <= k) return 0;
int left = 0, right = mx;
while (left < right) {
int mid = (left + right) >> 1;
ll t = 0;
for (int v : d) t += max(v - mid, 0);
if (t <= k)
right = mid;
else
left = mid + 1;
}
for (int i = 0; i < n; ++i) {
k -= max(0, d[i] - left);
d[i] = min(d[i], left);
}
for (int i = 0; i < n && k; ++i) {
if (d[i] == left) {
--k;
--d[i];
}
}
ll ans = 0;
for (int v : d) ans += 1ll * v * v;
return ans;
}
};

• func minSumSquareDiff(nums1 []int, nums2 []int, k1 int, k2 int) int64 {
k := k1 + k2
s, mx := 0, 0
n := len(nums1)
d := make([]int, n)
for i, v := range nums1 {
d[i] = abs(v - nums2[i])
s += d[i]
mx = max(mx, d[i])
}
if s <= k {
return 0
}
left, right := 0, mx
for left < right {
mid := (left + right) >> 1
t := 0
for _, v := range d {
t += max(v-mid, 0)
}
if t <= k {
right = mid
} else {
left = mid + 1
}
}
for i, v := range d {
k -= max(v-left, 0)
d[i] = min(v, left)
}
for i, v := range d {
if k <= 0 {
break
}
if v == left {
d[i]--
k--
}
}
ans := 0
for _, v := range d {
ans += v * v
}
return int64(ans)
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).