How to detect a common superview.

Question

Ganesh · Accepted Answer

- (nullable UIView *)commonSuperview:(nullable UIView *)otherView {
    if(!otherView)
        return nil;

    NSMutableArray* superViewsArray = [NSMutableArray array];
    UIView* superVeiw = self;

    while (superVeiw) {
        [superViewsArray addObject:superVeiw];
        superVeiw = superVeiw.superview;
    }

    superVeiw = otherView;
    while (superVeiw) {
        if([superViewsArray containsObject:superVeiw]){
            break;
        }
        superVeiw = superVeiw.superview;
    }
    return superVeiw;
}

Marcio · Answer

extension UIView {
    func sameSuperview(other: UIView) -> UIView? {
        guard self.superview != nil || other.superview != nil  else { return nil }
        
        var superViews: [UIView] = [] //Should be a set
        var selfSuperView = self.superview
        while selfSuperView != nil {
            superViews.append(selfSuperView!)
            selfSuperView = selfSuperView!.superview
        }
        selfSuperView = other.superview
        while selfSuperView != nil {
            if superViews.contains(selfSuperView!) {
                return selfSuperView
            }
            selfSuperView = selfSuperView!.superview
        }
        
        return nil
    }
}

Anonymous · Answer

- (UIView *)commonSuperView:(UIView *)oneView
                anotherView:(UIView *)anotherView {
    if (!oneView || !anotherView) {
        return nil;
    }
    NSMutableArray *viewArray = @[oneView].mutableCopy;
    while (oneView.superview) {
        oneView = oneView.superview;
        [viewArray addObject:oneView];
    }
    if ([viewArray containsObject:anotherView]) {
        return anotherView;
    }
    while (anotherView.superview) {
        anotherView = anotherView.superview;
        if ([viewArray containsObject:anotherView]) {
            return anotherView;
        }
    }
    return nil;
}

Anonymous · Answer

// The total complexity of the algorithm is O(h1) + O(h2)
    
    func commonView(view1:UIView,view2:UIView?) -> UIView?{
        
        if view2 != nil {
            var superView = [UIView:UIView]()
            var firstView = self.view
            var otherView = view2
            while(firstView != nil){
                
                superView[firstView!] = firstView?.superview ?? UIView()  // h1 is the hieght of view1 so the complexitfy here is O(h1)
                firstView = firstView?.superview
            }
            while(otherView != nil){
                //let say h1 is the height of view2  from its main parent. so the worst case complexity is O(h2)
                if (superView[otherView!] != nil){
                    return superView[otherView!]
                }
                otherView = otherView?.superview
            }
        }
        
        return nil
        
    }

Munir · Answer

This is like finding lca with parent link. 
Find depth of both views from root. Match level (Traverse upwards on longer chain). Technically you will be on same level as both root now just check parents,until they match go up the chain.

frank · Answer

- (UIView *)commonSuperview:(UIView *)otherView
{
    NSMutableSet *views = [NSMutableSet set];
    UIView *view = self;
    
    do {
        if (view != nil)
        {
            if ([views member:view])
            {
                return view;
            }
            [views addObject:view];
            view = view.superview;
            
            if (otherView != nil)
            {
                if ([views member:otherView])
                {
                    return otherView;
                }
                
                [views addObject:otherView];
                otherView = otherView.superview;
            }
        }
    } while (view || otherView);
    
    return nil;
}

This could work · Answer

func commonSuperView(view1: UIView, view2: UIView) -> UIView? {
    var tempView1: UIView =  view1
    var tempView2: UIView =  view2
    while tempView1.superview != nil {
        while tempView2.superview != nil {
            if tempView1.superview === tempView2.superview {
                return tempView1.superview!
            }
        }
        tempView1 = tempView1.superview!
        tempView2 = view2
    }
    return nil
}

Anonymous · Answer

func findCommonOne(view1: UIView, view2: UIView) -> UIView? {
        
        let tagConstant = 10000
        var view1 = view1
        var view2 = view2
        
        view1.tag = tagConstant
        while let parentView = view1.superview {
            parentView.tag = tagConstant
            view1 = parentView
        }
        
        while let parentView = view2.superview  {
            if parentView.tag == tagConstant {
                return parentView
            }
            view2 = parentView
        }
        
        return nil
    }

Anonymous · Answer

Simple unoptimized solution, followed by a solution using a hash map. Wasn't sure at first if it was possible to optimize.

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