Reduction Operations to Make the Array Elements Equal | Intuition | Leetcode-1887

This is the 68th Video on our Array Playlist
In this video we will try to solve a very good Array problem - Reduction Operations to Make the Array Elements Equal (Leetcode -1887).

I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.

Problem Name : Reduction Operations to Make the Array Elements Equal
Company Tags : Will update soon

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[DeepakSingh-fdix]

Sir, if you could include solutions of LeetCode contests, it would be greatly appreciated. Please add LeetCode solutions

[anshukmishra]

Consistency 🔥🔥🔥🔥. Bhiaya question solve karne ki aadat lagane ke liye aur hume guide karne ke liye thank you.

[anshukmishra]

class Solution {
public:
int nums) {
sort(nums.begin(), nums.end());
int count = 0;
int size = nums.size() - 1;
for(int i = size - 1 ; i >= 0; i--){
if(nums[i] == nums[i+1]){
continue;
}
count += (size-i);
nums[i+1] = nums[i];
}
return count;
}
};

[ammanchhetri]

My Logic - As we are decreasing the value...to be more precise...we are changing every element to its just smaller element...at the end...all the elements in the array will become equal to the minimum element of the array...so we just use a for loop and count how many operations it will take to make this particular element to minimum element....also the number of operations is equal to the number of unique values occured just before this number....so... :)




class Solution {
public:
int nums) {

int n = nums.size();

if(n == 1) return 0;

sort(nums.begin(), nums.end());

unordered_set<int> st;

int mini = nums[0];
int operations = 0;

for(int i = 0;i<n;i++) {
if(nums[i] == mini) {
continue;
}else if(nums[i] != nums[i-1]) {
st.insert(nums[i-1]);
operations += st.size();
}else{
operations += st.size();
}
}


return operations;

}
};

[DevOpskagyaan]

Leetcode 2939 - whenever you get time. No rush. Enjoy your Sunday

[creatorstudio]

Great it by myself after watching half video.

[mewaconite]

Great explanation... as always... thank you for making this video :)
ye wala khud se hi ban gya tha ....
the intuition that came to my mind after spending 3-5 mins was sort and count, ... after watching your video... i learned we could do reverse loop and then add (n - i)

class Solution {
public:
int nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
int answer = 0;
int count = 0;
int previous = nums[0];

for (int i=0; i< n; i++){
if (previous != nums[i]){
count++;
}
answer += count;
previous = nums[i];
}
return answer;
}
};

[sauravchandra]

My approach:

int ans=0, n=nums.size(), i=1, cnt=0;

sort(begin(nums), end(nums));

while(i<n && nums[i-1]==nums[i]) i+=1;

for(i;i<n;){
int j=i;
cnt+=1;
while(j<n && nums[j]==nums[i]) j+=1;
ans += (j-i)*cnt;
i=j;
}

return ans;

[mylearning]

class Solution {
public:

static bool comp(int a, int b){
return b<a;
}

int nums) {
sort(nums.begin(), nums.end(), comp);
int n=nums.size(), i=0, j=1, ans=0;
while(n>1 && i<j && j<n){
if(nums[i]>nums[j]){
nums[i]=nums[j];
ans+=(j-i);
}else if(nums[i]==nums[j]){
j++;
}
}
return ans;

}
};

[DurgaShiva]

Hi All
Apart from the most optimal and amazing approach shared by code story with mik, I have also tried one more way to solve the same problem and the code is as follows : -
class Solution {
public int reductionOperations(int[] nums) {
TreeMap<Integer, Integer> tm = new
for(int i=0;i<nums.length;i++)
{
tm.put(nums[i], tm.getOrDefault(nums[i], 0)+1);
}

int biggest = tm.firstKey();
int smallest = tm.lastKey();
int prev = 0;
int curr = 0;

while(biggest != smallest)
{
curr = tm.get(biggest);
prev+=curr;
tm.remove(biggest);
biggest = tm.firstKey();
tm.put(biggest, tm.get(biggest)+curr);

}
return prev;
}
}

[saurabhKumar-hjyp]

I also solved the question with the same approach before watching the video. As you told in the video .

[lofireverbz-wygo]

hey you! yes you mik ji,
thanks for making video like this 🥰

[pranjalyadav]

broo contest kae k solutions bhi bnayae kriyaee .
your explaination good enough

[chiragsingh]

Make video on today's 3rd question of contest.

[bobo_v]

Hello bhaiyaa can you mention your time to solve that particular question because sometimes i feel ki mujhe itna time lag rha easy questions pe bhi ye comparison ke liye nhi apko mujhse bahot jyda ata hai ap to mentor jaise ho pr sirf generally time mention kr dena ki kitna time mai apse solve huaa video wala question even ani wali sari video mai

[infinitygaming]

nums.sort()
p = 0
k = 0
for i in range(len(nums)-1):
if nums[i] != nums[i+1]:
k += 1
p += k

return p

[FanIQQuiz]

World Cup k baad contest ka last Qn dekhna bhaiya please 😄

[Ramneet]

Bhaiya even i did thiss solution by own, but people have done with o(n)
By taking space like by using hash array and having freq and i didn't get that approach, can you please make a vedio on that, even i think yesterday or day before yesterday we got a question like that where we did sorting but on discuss form many have done with o(n).
Thx

[saumilvachheta]

Hi, can you record the videos in screen ratio same as other YouTube videos? Just a suggestion.

[S__Arslaan]

Reverse sort karke aur bhi easy lagta h. ops mei i+1 add krna h