870. Advantage Shuffle
Medium
TopicsCompanies
You are given two integer arrays nums1 and nums2 both of the same length. The advantage of nums1 with respect to nums2 is the number of indices i for which nums1[i] > nums2[i].
Return any permutation of nums1 that maximizes its advantage with respect to nums2.
 
Example 1:
Input: nums1 = [2,7,11,15], nums2 = [1,10,4,11]
Output: [2,11,7,15]


Example 2:
Input: nums1 = [12,24,8,32], nums2 = [13,25,32,11]
Output: [24,32,8,12]


 
Constraints:
1 <= nums1.length <= 105
nums2.length == nums1.length
0 <= nums1[i], nums2[i] <= 109

Bahii 3ana problem ismo “Advantage Shuffle” oky : 
—> Fe problem h4a 3ana two lists “nums1” and “nums2” el zouz 3andhom  nefess el toull … 
—> oky bsh t7eb “the advantage” of list x relatively 3le y lzem , el indices “i” ykonn 
—> nums1[i] > nums2[i] for kol indice “i” oky 

Bahii el problem honi y7ebik tl3 any permutation fe “nums1” t5alik “maximize” it advantage % 3le nums2 

Q- awl seoul chnowa el input & output format ? 

—> input : nums1:List[int] , nums2:List[int] 
—> output : permuted_nums1:List[int] 


Q- oky first question did i can also shuffle num2 yes or non ? 

Non 
Q- oky 




In this example : nums1=[2,7,11,15] , nums2=[1,10,4,11] 


In this example (without any shuffling) 

That advantage rank of nums1 % 3le nums2 is 3 kefh ? 

nums1[0]>nums2[0] —> +1
Nums1[2] >nums2[2] —>+1
Nums1[3] > nums3[3] —>+1

Q- ame z3ma ha4ii e7sen “configuration” of nums1 possible ? bsh we maximize the advantage rank? 

Q- ntswer lee ? tnjm t3tena configuration 5er ? 
Ken we souffle nums1[1]<->nums1[2] we get 

A rank = 4 >3 oky donc njmo nl9wa 5er 

Q- twa ha4a logic 5aterik entik 3andikk “eyes” and percetront 5aletik fe example sghir kema hka 
Treason about a better configuration oky : ame z3ma njmo “generalize” a better a more format and systematic abrouch , be m3na h4a njmo n5l9o algorithm bien verfier bsh ysolvy this problem ? 


Q- n7wloo mele ? ntswer fema paradigm mt3 problem solving “ynjm” y9rabna el solution ideal enou enik entii ? twell fe kol mara tmaximizy the solution local (tw b3ed n3mlo definition leha) 3le ame enik tl9a solution global bien confirmer oky ? 

Q- trh kefhhh ? oky 

Q- awl 7aje 5alina n7wloo we define the local optimum solution  in this problem ? 

—> bahi honi the local optimum howa el enik tal9a “i” where nums1[i] >nums2[i] 
Fel position ha4ii 

Q- donc kefh? Njmo we granted this local optimum condition ? 

→ i think we have two step (precompute step and on fly solution oky ) 

The precompute step this first sort the input list “nums1” in a non-decreasing order oky 


→ the second step : is to use on fly pointer updation to update the value of nums1[i] if we get 
nums1[i]<nums[j] 

Note the actual logic of how we efficiently update the value need more investigation oky 
→ but for now we stop 
Q- we have done a lot  of work and thinking process so far ? 
Q- this why i need some guidance 
Q- first to know if am i in the right direction yes or non 
Q- second to help me reduce the search space and reduce my energie to focus on the most important chanallges oky
Q- do not guive actual solution with code , just more abstract ideas and hints 
Q- be my mentor here oky 

Let me help guide your thinking process for the Advantage Shuffle problem. You've already done some good analysis!

Let's break down what we know and provide some direction:

1. Your initial analysis is correct:
- We can only shuffle nums1
- We need to maximize the count of positions where nums1[i] > nums2[i]
- Local optimization is about finding good matchups

2. Your intuition about sorting nums1 is a great start! Here's why:
- When we have sorted nums1, we have more control over which value we want to use against each nums2[i]
- This gives us the flexibility to make optimal "matchups"

Here's a key insight to consider:
Think of this as a matching game. For each position in nums2, we want to:
- Either beat it with the smallest possible number from nums1 that's larger (why waste a bigger number if a smaller one can win?)
- Or if we can't beat it, what should we do with our current number?

Some guiding questions to help you develop the algorithm:
1. If we sort nums1, how can we efficiently find the smallest number that beats a given nums2[i]?
2. What should we do with numbers from nums1 that can't beat any number in nums2?
3. How can we keep track of which numbers from nums1 we've already used?

You're definitely on the right track with sorting nums1. The next step is to think about:
- How to efficiently match numbers from sorted nums1 to nums2
- What to do with "leftover" numbers that can't win any comparison

Would you like to think about any of these aspects in more detail?

Remember: The goal is to be greedy but smart about which numbers we use to beat each position in nums2.


Q- oky twa lzemna nchof el logic mt3 mapping eli ynjm ykon honii ? oky 

Q- awll 7aje jet le mo5ii enou nts3mloo two pointers (i and j) 

i just iterate from left to right in the sorted_nums1 (from the smallest one to the biggest ) 

We j zeda men left to right leken honii el m3nch el logic keml est natively men left to right 
Wle fema more sophisticated technique 


Q- 5alina we proof the correctness of the last method using this example 

nums1 = [12,24,8,32], nums2 = [13,25,32,11]

Bahii nums1.sort() t3tina [8,12,24,32] & nums2=[13,25,32,11] 

Hanii ntswer el naive logic mch s7i7 barsha 

Ntswer lzmna data structure t5alina nextract el max value mt3 nums2 kol mara (m3 index mt3ha ) and we compare it with the min value of nums2 

Ken akbar integer fe nums2 asghir men ‘asghir” elemnt fe nums2 
→ donc directl the sorted list is the best configuration to get the max rank oky 


Q- we have done a lot  of work and thinking process so far ? 
Q- this why i need some guidance 
Q- first to know if am i in the right direction yes or non 
Q- second to help me reduce the search space and reduce my energie to focus on the most important chanallges oky
Q- do not guive actual solution with code , just more abstract ideas and hints 
Q- be my mentor here oky 


Let me help guide your thought process further. You're making good progress with considering different mapping strategies!

Let's analyze your latest thoughts:

1. Two Pointers Approach:
- Your intuition about using two pointers is good, but you correctly realized that a simple left-to-right scanning might not be optimal
- Why? Because we might waste smaller numbers on larger numbers in nums2 when we could have used them more effectively elsewhere

2. Your insight about needing to track maximum values in nums2 is interesting!
Let's think about this differently:

Consider this strategy:
1. For each number in nums2, what if we:
   - Keep nums2 in original order but track their indices
   - Sort nums1 (you already got this part!)
   - For each value in nums2, try to find the smallest number in sorted nums1 that can beat it
   - If we can't beat it, what's the optimal thing to do with our smallest available number?

Key Questions to Consider:
1. Would using a data structure like a binary search tree or priority queue help us efficiently find:
   - The smallest number in nums1 that's greater than current nums2[i]?
   - Track which numbers we've already used?

2. For the example: nums1 = [8,12,24,32], nums2 = [13,25,32,11]
- What happens if we can't beat 13 with 8?
- Should we "waste" 8 here or save it for 11 which appears later?
- How do we efficiently know that 11 appears later?

Think about:
- Could sorting both arrays (while keeping track of original indices) help?
- What's the optimal strategy when we can't beat a number?
- Is there a way to process numbers in a more strategic order than left-to-right?

Would you like to explore any of these aspects further?


Q- now let’s to apply many concepts here and see how the logic is done ? 

First : ntesewer taw n7na mteked enou sorting el input list “nums1” → yukon a good starting point oky 
 nums1 = [8,12,24,32]


Second : netwser lzemna mechnisime y5alina nt5rfoo nmapping , and compare coll mar asghir and akbar element here : 

Bahii el logic bsh yukon kema hka : n7na mnech bsh “suffle the nums2” bel bel3kess bsh n7wloo we maintain the idex of their original presence laken bsh n7wloo we sort the value “value” of el element eli idex howa ypointy 3lih oky
Q- do el soule kefh bsh ykon el mecanisme honii bsh t3ml hka : 

Netwer bsh n7wloo we change list[int] → twlii list[Tuple[value,index]] 
Then we sort it based on the value oky 

For example  nums2 = [13,25,32,11]
nums2_tuple=[[13,0],[25,1],[32,2],[11,3]] 
—> then we sort it based on it’s value 
[[11,3],[13,0],[25,1],[32,2]] okt 

Q- twa 3ana ? honi 

  [8,12,24,32] , [[11,3],[13,0],[25,1],[32,2]] 


Q- twa ntswer lzemna we suffle the input array based on the goal of “maximsing the value “

// first we need to initiate a left pointer
left=0
// second we need to initiate a right pointer 
right=nums1.length() 

Then we need to traverse the second modified list from the largest to smallest : 

Q- bahii twa ntswer bdena n9robo lel solution final ? : 
Q- tnjm as mentor t5lik diusccsion (active between you and me) and
The all the options to  n7wlo nzedo n9ribo ? 

Q1: In your current setup, why do you want to traverse the modified nums2 from largest to smallest? What advantage does this give us?

—> ntswer 3lh n7well i trzverse men a9bel lel asghir t5alini napplique the principle of greedy algorithm 
→ enik tjwebb “ 3lehh nzedd not5ol fe details ken 9ett enou akber 3ded fe nums2 how asghir men a9bell 3ed fe nums1 “ donc the update mt3 suffling bsh yukon sahel 
Just n5alik fe balsto “ and ken lel just we go through the next greatest one oky 

Q- el jweb honi abhii ey wle ?

Q- oky 5alina nt7wloo nzedo ? 

Bahii 5alina ntsweroo enou 9ina : 

nums1 // si deja sorted 
Nums2 // wlet list of tuple men akbar element le asghir w7ed oky 

// 5alina we initate two pointers left and right oky 
left=0
Right = n-1
// twa 5alina n7wllo we iterate through the linst nums2 oky 
For i in range(n):
	// bahi kena enou a9bell element fe nums1 how akber men akber element from nums2
	If nums1[right] >nums2[i[0]]: 
			// just we assign the value nums1[right] fe index nums2[i[1]]:
			idx=nums2[i[1]]
			nums1[idx]=nums1[right] 
			// just move to the n-1 largest on 
			right-=1
	Else : 
			// just we assign the value nums1[left] le index nums2[i]: 
			// souelly twa 3lehhh kha else docntion and twnjm t3ml example bsett bsh twrini 3leh