1509. Minimum Difference Between Largest and Smallest Value in Three Moves Leetcode daily challenge
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- Опубликовано: 1 июл 2024
- Problem:
1509. Minimum Difference Between Largest and Smallest Value in Three
Problem Statement:
You are given an integer array nums.
In one move, you can choose one element of nums and change it to any value.
Return the minimum difference between the largest and smallest value of nums after performing at most three moves.
Problem Link:
leetcode.com/problems/minimum...
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Timestamp:
0:00 - Introduction
#ShashwatTiwari #coding #problemsolving
To be very honeest very well and pin point explanation
Thanks 😊
very nice and neat explaination , thank you buddy 👍👍
Beautiful explanation , i always look into your solution and become able to write code on my own ,
kal channel mila aur aaj se daily problem solving continue pura month solve karunga
You explains really well bhaiya.
Super easy solution, Thanks
Very good explanation. Ek bachhe ko bhi samajh aa jayega Aisa explanation. ❤
brilliant logic!!
Yrr Awesome explanation brother always waiting for your video even after solving the problem by own ❤
hats off for this explanation!!
please continue this series love from canada🥰
Amazing explanation sir
good explain sir
Osm explain sir
class Solution {
public int minDifference(int[] nums) {
int n = nums.length; // Step 1: Get the length of the array.
if (n
Single loop looks neat!👏👌
cool
3:10 🤣🤣🤣
without using sorting i get the four minimum and four maximum values
class Solution {
public int minDifference(int[] nums) {
if (nums.length < 5) {
return 0;
}
// Initialize the four smallest and four largest values
int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE, min3 = Integer.MAX_VALUE, min4 = Integer.MAX_VALUE;
int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE, max4 = Integer.MIN_VALUE;
for (int num : nums) {
// Update the four smallest values
if (num < min1) {
min4 = min3;
min3 = min2;
min2 = min1;
min1 = num;
} else if (num < min2) {
min4 = min3;
min3 = min2;
min2 = num;
} else if (num < min3) {
min4 = min3;
min3 = num;
} else if (num < min4) {
min4 = num;
}
// Update the four largest values
if (num > max1) {
max4 = max3;
max3 = max2;
max2 = max1;
max1 = num;
} else if (num > max2) {
max4 = max3;
max3 = max2;
max2 = num;
} else if (num > max3) {
max4 = max3;
max3 = num;
} else if (num > max4) {
max4 = num;
}
}
// Calculate the minimum difference after removing up to 3 elements
int a = max4 - min1;
int b = max3 - min2;
int c = max2 - min3;
int d = max1 - min4;
return Math.min(Math.min(a, b), Math.min(c, d));
}
}
dudee...plzz plz plzzzz roz ke dcc ke solutions dalna shuru kar do yaarr/...bahout sexy explanation mann
3:09 💀
Bhaiya, please make a video on Leetcode 2035 problem. I'm stuck on this problem since 2 weeks.
so after sorting we only need to change the element size by there next value??
Noise
How to prove that there are only 4 cases ?
its simple combinatorics as we can change only 3 positions and we can only choose the positions from the begin and the end. now the distributions possible are -> 3 from front 0 from last, 2 from front 1 from last,1 from front 2 from last , 0 from front and 3 from last. you can't make any other possible distribution.