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Hey , Can you please Explain your solution with help of some real life analogy or any intuition how you ended up getting this sloution ??
I did this by eliminating the no which didn't occur in the result of one window , I guess its O( nlog(n) ) ...
For Example :
You have { 2, 6 ,1, 12 }
After First Pass we get : { 2 , 6 , 1, 12} , Max = 12;
After Second Pass : { 2, 1, 1} , Max = 2;
Now what I will be doing is only comparing the results obatined in second pass rather than the original array . So Third Pass would be obtained by Taking mins of ( 2,1) & ( 1,1 ) & finding out the max which will be 1 .
Third Pass : { 1,1 } , Max =1 ;
Fourth Pass : {1} , Max = 1;
Can you please elaborate your approach intuition ??
Min Max Riddle
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Hey , Can you please Explain your solution with help of some real life analogy or any intuition how you ended up getting this sloution ??
I did this by eliminating the no which didn't occur in the result of one window , I guess its O( nlog(n) ) ... For Example :
You have { 2, 6 ,1, 12 }
After First Pass we get : { 2 , 6 , 1, 12} , Max = 12; After Second Pass : { 2, 1, 1} , Max = 2;
Now what I will be doing is only comparing the results obatined in second pass rather than the original array . So Third Pass would be obtained by Taking mins of ( 2,1) & ( 1,1 ) & finding out the max which will be 1 .
Third Pass : { 1,1 } , Max =1 ; Fourth Pass : {1} , Max = 1;
Can you please elaborate your approach intuition ??